Question 1

Question 51300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20° C to 40° C. Calculate the specific heat capacity of lead.

Accepted Answer

Given,Heat energy supplied (Q) = 1300 JMass of lead (m) = 0.5 kgChange in temperature (△t) = (40 – 20)°C = 20° CSpecific heat capacity (c) = ?From relation,Q=m×c×△tQ = m \times c \times △t \[0.5em]Q=m×c×△tSubstituting the values in the formula above we get,1300=0.5×c×20⇒c=13000.5×20⇒c=130 J kg−1K−11300 = 0.5 \times c \times 20 \[0.5em] \Rightarrow c = \dfrac{1300}{0.5 \times 20} \[0.5em] \Rightarrow c = 130 \text{ J kg}^{-1} \text{K}^{-1}1300=0.5×c×20⇒c=0.5×201300​⇒c=130 J kg−1K−1Hence, specific heat capacity of lead = 130 J kg-1 K-1

Question 2

Question 6A car's cooling system uses water to absorb heat from the engine. If 5 kg of water absorbs 420 kJ of heat, what is the temperature increase of water? (Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹)

Accepted Answer

Given,Mass of water (m) = 5 kgHeat absorbed (Q) = 420 kJ = 420×103420	imes10^3420×103 JSpecific heat capacity of water (c) = 4200 J kg⁻¹ K⁻¹Let, increase in temperature be △t.From relation,Q=mc△tQ = m c △tQ=mc△tOn rearranging terms,△t=Qmc△t = \dfrac{Q}{mc}△t=mcQ​Substituting the values in the formula above we get,△t=420×1035×4200=1005=20K=20 oC△t = \dfrac{420	imes10^3}{5	imes4200}=\dfrac{100}{5}=20 K= 20\ ^oC△t=5×4200420×103​=5100​=20K=20 oCTemperature of water will increase by 20 °C.

Question 3

Question 9An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0° C to 15.0° C in 100 s. Calculate — (i) the heat capacity of 4.0 kg of liquid, and (ii) the specific heat capacity of liquid.

Accepted Answer

(i) Power of heater (P) = 600 WMass of liquid (m) = 4.0 kgChange in temperature of liquid = (15 – 10)°C = 5° C (or 5 K)Time taken to raise it's temperature (t) = 100 sheat capacity = ?From relation,Q=Power×timeQ = \text {Power} \times \text{time}Q=Power×timeSubstituting the values in the formula above we get,Q=600×100⇒Q=60000JQ = 600 \times 100 \[0.5em] \Rightarrow Q = 60000 J \[0.5em]Q=600×100⇒Q=60000JNow,C′=Q△TC^{'} = \dfrac{Q}{△T} \[0.5em]C′=△TQ​Substituting the values in the formula above we get,C′=60,0005C′=1.2×104 JK−1C^{'} = \dfrac{60,000}{5} \[0.5em] C^{'} = 1.2 \times 10^{4} \text{ J} \text{K}^{-1} \[0.5em]C′=560,000​C′=1.2×104 JK−1Hence, heat capacity = 1.2 x 10 J K-1(ii) specific heat capacity {c} = ?c=Qm×△Tc = \dfrac{Q}{m \times △T} \[0.5em]c=m×△TQ​Substituting the values in the formula above we get,c=600004×5c=600004×5c=3×103 J kg−1K−1c = \dfrac{60000}{4 \times 5} \[0.5em] c = \dfrac{60000}{4 \times 5} \[0.5em] c = 3 \times 10^{3} \text{ J kg}^{-1} \text{K}^{-1}c=4×560000​c=4×560000​c=3×103 J kg−1K−1Hence, specific heat capacity = 3 x 103 J Kg-1 K-1

Question 4

Question 1245 g of water at 50°C in a beaker is cooled when 50 g of copper at 18°C is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39 J g-1 K-1 and that of water is 4.2 J g-1 K-1. State the assumptions used.

Accepted Answer

Given,Mass of water = 45 gLet the final constant temperature reached be t°CFall in temperature of water = (50 – t)°CMass of copper = 50 gRise in temperature of copper = (t - 18)°CThe specific heat capacity of the copper cc = 0.39 J g-1 K-1The specific heat capacity of water cw = 4.2 J g-1 K-1Heat energy given by water = mc△t= 45 x 4.2 x (50 – t)     [Equation 1]Heat energy taken by copper = 50 x 0.39 x (t - 18)     [Equation 2]Assuming that there is no loss of heat energyHeat energy given by water = Heat energy taken by copperEquating equations 1 & 2, we get,45×4.2×(50−t)=50×0.39×(t−18)⇒189×(50−t)=19.5×(t−18)⇒(189×50)−(189×t)=(19.5×t)−(19.5×18)⇒(9450)−(189×t)]=(19.5×t)−(351)⇒9450+351=(19.5t)+(189t)⇒9801=(208.5t)⇒t=9801208.5⇒t=47.0072°C≈47°C45 \times 4.2 \times (50 - t) = 50 \times 0.39 \times (t - 18) \[0.5em] \Rightarrow 189 \times (50 - t) = 19.5 \times (t - 18) \[0.5em] \Rightarrow (189 \times 50) - (189 \times t) = (19.5 \times t) - (19.5 \times 18) \[0.5em] \Rightarrow (9450) - (189 \times t)] = (19.5 \times t) - (351) \[0.5em] \Rightarrow 9450 + 351 = (19.5 t) + (189 t) \[0.5em] \Rightarrow 9801 = (208.5t) \[0.5em] \Rightarrow t = \dfrac{9801}{208.5} \[0.5em] \Rightarrow t = 47.0072°C \approx 47°C45×4.2×(50−t)=50×0.39×(t−18)⇒189×(50−t)=19.5×(t−18)⇒(189×50)−(189×t)=(19.5×t)−(19.5×18)⇒(9450)−(189×t)]=(19.5×t)−(351)⇒9450+351=(19.5t)+(189t)⇒9801=(208.5t)⇒t=208.59801​⇒t=47.0072°C≈47°CHence, final temperature = 47°C

Question 5

Question 13200 g of hot water at 80°C is added to 400 g of cold water at 10°C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg-1 K-1.

Accepted Answer

Mass of hot water = 200 gTemperature of hot water = 80° CMass of cold water = 400 gTemperature of cold water = 10° CLet final temperature be tFall in temperature of hot water = (80 – t)°CRise in temperature of cold water = (t - 10)°CThe specific heat capacity of water cw = 4200 J kg-1 K-1 = 4.2 J g-1 K-1Heat energy given by hot water = mc△t= 200 x 4.2 x (80 – t)     [Equation 1]Heat energy taken by cold water = 400 x 4.2 x (t - 10)     [Equation 2]Assuming that there is no loss of heat energy,Heat energy given by hot water = Heat energy taken by cold waterEquating equations 1 & 2, we get,200×4.2×(80−t)=400×4.2×(t−10)2×(80−t)=4×(t−10)160−2t=4t−40160+40=4t+2t200=6t⇒t=2006⇒t=33.3°C200 \times 4.2 \times (80 - t) = 400 \times 4.2 \times (t - 10) \[0.5em] 2 \times (80 - t) = 4\times (t - 10) \[0.5em] 160 - 2t = 4t - 40 \[0.5em] 160 + 40 = 4t + 2t \[0.5em] \[0.5em] 200 = 6t \[0.5em] \Rightarrow t = \dfrac{200}{6} \[0.5em] \Rightarrow t = 33.3°C200×4.2×(80−t)=400×4.2×(t−10)2×(80−t)=4×(t−10)160−2t=4t−40160+40=4t+2t200=6t⇒t=6200​⇒t=33.3°CHence, final temperature = 33.3°C

Question 6

Question 14Two blocks P and Q of different metals having their mass in the ratio 2 : 1 are given same amount of heat. Their temperature rises by same amount. Compare their specific heat capacities.

Accepted Answer

Let,Specific heat capacity of block P = CpSpecific heat capacity of block Q = CQFrom relation,c=Qm×△tc = \dfrac{Q}{m \times △t} \[0.5em]c=m×△tQ​where, c = specific heat capacitym = massQ = heat energy△t = change in temperatureNow,CpCQ=Q2m×△tQm×△tCpCQ=m×△t2m×△t⇒CpCQ=12\dfrac{C_p}{C_Q} = \dfrac{\dfrac{Q}{2m × △t}}{\dfrac{Q}{m × △t}} \[0.5em] \dfrac{C_p}{C_Q} = \dfrac{m × △t}{2m × △t} \[0.5em] \Rightarrow \dfrac{C_p}{C_Q} = \dfrac{1}{2} \[0.5em]CQ​Cp​​=m×△tQ​2m×△tQ​​CQ​Cp​​=2m×△tm×△t​⇒CQ​Cp​​=21​Hence, the required ratio is 1 : 2

Question 7

Question 19Why the base of a cooking pan made thick and heavy?

Accepted Answer

The base of a cooking pan is made thick and heavy because it's heat capacity becomes large due to which it gets heated slowly and it imparts sufficient heat energy at a slow rate to the food for it's proper cooking and after cooking it keeps the food warm for a long time.

Question 8

Question 2How is the heat capacity of a body related to the specific heat capacity of it's substance?

Accepted Answer

The equation which relates the heat capacity of a body to the specific heat capacity is —C=m×cC=m	imes cC=m×c

Question 9

Question 120 g of ice at 0° C absorbs 10,920 J of heat energy to melt and change to water at 50° C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg-1 K-1.

Accepted Answer

Given,Mass (m) = 20 gHeat energy absorbed (Q) = 10,920 JSpecific heat capacity of water = 4200 J kg-1 K-1 = 4.2 J g-1 K-1Specific latent heat of fusion of ice (L) = ?(i) Heat energy required to melt the ice at 0° C to water at 0° C (Q1) = m x LSubstituting the values in the formula we get,Q1=20×LQ_1 = 20 \times LQ1​=20×L(ii) Heat energy required to raise temperature from 0° C to 50° C = m x c x rise in temperatureSubstituting the values in the formula we get,Q2=20×4.2×50=4200 JQ_2 = 20 \times 4.2 \times 50 \[0.5em] = 4200 \text{ J}Q2​=20×4.2×50=4200 JFrom relation,Q=Q1+Q210,920=(20× L)+420010,920−4200=20×L6720=20×L⇒L=672020⇒L=336 J g−1Q = Q_1 + Q_2 \[0.5em] 10,920 = (20 × \text{ L}) + 4200 \[0.5em] 10,920 - 4200 = 20 × L\[0.5em] 6720 = 20 × L\[0.5em] \Rightarrow L = \dfrac{6720}{20} \[0.5em] \Rightarrow L = 336 \text{ J g}^{-1} \[0.5em]Q=Q1​+Q2​10,920=(20× L)+420010,920−4200=20×L6720=20×L⇒L=206720​⇒L=336 J g−1Hence, Specific latent heat of fusion of ice = 336 J g-1

Question 10

Question 3A molten metal of mass 150 g is kept at it's melting point 800° C. When it is allowed to freeze at the same temperature, it gives out 75,000 J of heat energy.(a) What is the specific latent heat of the metal?(b) If the specific heat capacity of metal is 200 J kg-1 K-1, how much additional heat energy will the metal give out in cooling to - 50° C?

Accepted Answer

(a) Given,Mass (m) = 150 gHeat energy given out (Q) = 75,000 JSpecific latent heat of the metal = ?From relation Q = m x LSubstituting the values in the formula we get,75000=150×L⇒L=75000150⇒L=500 J g−175000 = 150 \times L \[0.5em] \Rightarrow L = \dfrac{75000}{150} \[0.5em] \Rightarrow L = 500 \text{ J g}^{-1} \[0.5em]75000=150×L⇒L=15075000​⇒L=500 J g−1Hence, the specific latent heat of the metal = 500 J g-1(b) Specific heat capacity of metal is 200 J kg-1 K-1Change in temperature= 800 – (-50) = 800 + 50 = 850° C = 850 KFrom relation,Q = m x c x change in temperatureSubstituting the values we get,Q=0.15×200×850Q=25,500JQ = 0.15 \times 200 \times 850 \[0.5em] Q = 25,500 J \[0.5em]Q=0.15×200×850Q=25,500JHence, 25,500 J of heat energy will be given out

Question 11

Question 4A solid metal of mass 150 g melts at it's melting point of 800° C by providing heat at the rate of 100 W. The time taken for it to completely melt at the same temperature is 4 min. What is the specific latent heat of fusion of the metal?

Accepted Answer

Given,m = 150 g = 0.15 kgP = 100 Wt = 4 min = 240 sspecific latent heat of fusion of the metal = ?Heat supplied = P x tSubstituting the values we get,Heat supplied=100×240Heat supplied=24000J\text{Heat supplied} = 100 \times 240 \[0.5em] \text{Heat supplied} = 24000 JHeat supplied=100×240Heat supplied=24000JWe know,Q = m x LSubstituting the values we get,24000=0.15×L⇒L=240000.15⇒L=160,000⇒L=1.6×105 J kg−124000 = 0.15 \times \text{L} \[0.5em] \Rightarrow \text{L} = \dfrac{24000}{0.15} \[0.5em] \Rightarrow \text{L} = 160,000 \[0.5em] \Rightarrow \text{L} = 1.6 \times 10^{5} \text{ J kg}^{-1}24000=0.15×L⇒L=0.1524000​⇒L=160,000⇒L=1.6×105 J kg−1Hence, specific latent heat of fusion of the metal = 1.6 x 105 J kg-1

Question 12

Question 7The temperature of 170 g of water at 50° C is lowered to 5° C by adding certain amount of ice to it. Find the mass of ice added. Given : specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of ice = 336000 J kg-1.

Accepted Answer

Given,mw = 170 g = 0.17 kgspecific heat capacity of water = 4200 J kg-1 K-1specific latent heat of ice = 336000 J kg-1mi = ?Heat energy given out by water in lowering it's temperature from 50° C to 5° C= m x c x change in temperature= 0.17 x 4200 x (50 - 5)= 0.17 x 4200 x 45= 32,130Heat energy taken by m kg ice to melt into water at 0° C= mi x L= mi x 336000Heat energy taken by water at 0° C to raise it's temperature to 5° C= mi x c x change in temperature= mi x 4200 x (5 - 0)= mi x 4200 x 5= mi x 21000heat energy released = heat energy takenSubstituting the values we get,32130=(mi×336000)+(mi×21000)32130=mi×357000⇒mi=32130357000⇒mi=0.09 Kg=90 g32130 = (m_i \times 336000 ) + (m_i \times 21000 ) \[0.5em] 32130 = m_i \times 357000 \[0.5em] \Rightarrow m_i = \dfrac{32130}{357000} \[0.5em] \Rightarrow m_i = 0.09 \text{ Kg} = 90 \text{ g}32130=(mi​×336000)+(mi​×21000)32130=mi​×357000⇒mi​=35700032130​⇒mi​=0.09 Kg=90 gHence, the mass of ice added = 90 g

Question 13

Question 8Find the result of mixing 10 g of ice at -10° C with 10 g of water at 10° C. Specific heat capacity of ice = 2.1 J g-1 K-1, specific latent heat of ice = 336 J g-1, and specific heat capacity of water = 4.2 J g-1 K-1.

Accepted Answer

Given,mass of ice = 10 gmass of water = 10 gSpecific heat capacity of ice = 2.1 J g-1 K-1specific latent heat of ice = 336 J g-1,specific heat capacity of water = 4.2 J g-1 K-1Let the final temperature be t° C.Heat energy taken by ice at – 10° C to raise it's temperature to 0° C (Q1)= m x c x change in temperature= 10 x 2.1 x [0 - (-10)]= 10 x 2.1 x 10= 210 JHence, Q1 = 210 JHeat energy taken by ice at 0° C to convert into water at 0° C (Q2)= m x Lice= 10 × 336= 3360 JHence, Q2 = 3360 JHeat energy taken by water at 0° C to raise it's temperature to t° C (Q3) = m x c x change in temperature= 10 × 4.2 × (t – 0)= 10 × 4.2 × t= 42 tHence, Q3 = 42 tHeat energy released by water at 10° C to lower it's temperature to t° C (Q4) = m x cwater x change in temperature= 10 × 4.2 × (10 – t)= 42 × (10 – t) = 420 – 42tHence, Q4 = 420 – 42tIf there is no loss of heat,Heat energy gained = Heat energy lost210+3360+42t=420−42t3570−420=−42t−42t3150=−84tt=−315084t=−37.5°C210 + 3360 + 42 t = 420 - 42t \[0.5em] 3570 - 420 = - 42 t - 42 t \[0.5em] 3150 = - 84 t \[0.5em] t = - \dfrac{3150}{84} \[0.5em] t = - 37.5° C210+3360+42t=420−42t3570−420=−42t−42t3150=−84tt=−843150​t=−37.5°CThis cannot be true because water cannot exist at -37.5° C.Hence, we can say that the whole of the ice did not melt.Let amount of ice which melts = m gFinal temperature of the mixture = 0° CHeat energy gained by ice at -10° C to raise it's temperature to 0° C= m x cice x change in temperature= 10 × 2.1 x [0 - (-10)]= 10 × 2.1 x 10= 210 JHeat energy gained by m gm of ice at 0° C to change into water at 0° C= m × Lice= m x 336= 336m JHeat energy released by 10 g of water at 10° C to lower it's temperature to 0° C= m x cwater x change in temperature= 10 × 4.2 × (10 – 0)= 10 × 4.2 × 10 = 420 JIf there is no loss of heat,Heat energy gained = Heat energy lost210+336m=420⇒336m=420−210⇒336m=210⇒m=210336⇒m=0.625g210 + 336 m = 420 \[0.5em] \Rightarrow 336 m = 420 - 210 \[0.5em] \Rightarrow 336 m = 210 \[0.5em] \Rightarrow m = \dfrac{210}{336} \[0.5em] \Rightarrow m = 0.625 g210+336m=420⇒336m=420−210⇒336m=210⇒m=336210​⇒m=0.625gHence, 0.625 g of ice will melt and temperature will remain at 0° C

Question 14

Question 10Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C. Specific heat capacity of calorimeter = 0.4 J g-1 °C-1, Specific heat capacity of water = 4.2 J g-1 °C-1, Latent heat capacity of ice = 330 J g-1.

Accepted Answer

Given,mass of water mw = 150 gmass of calorimeter mc = 50 gSpecific heat capacity of calorimeter = 0.4 J g-1 °C-1,Specific heat capacity of water = 4.2 J-1 °C-1,Latent heat capacity of ice = 330 J g-1mass of ice mi = ?Heat energy imparted by calorimeter and water contained in it in cooling from 32° C to 5° C is used in melting ice and then raising the temperature of melted ice from 0°C to 5°C.Heat energy imparted by water (Q1)= m x c x change in temperature= 150 x 4.2 x (32 - 5)= 150 x 4.2 x 27= 17,010 JHeat energy imparted by calorimeter (Q2)= m x c x change in temperature= 50 x 0.4 x (32 - 5)= 50 x 0.4 x 27= 540 JHeat energy taken by ice to melt (Q3)= mi x L= mi x 330Heat energy gained by water from melted ice to reach from 0°C to 5°C (Q4)= mi x c x change in temperature= mi x 4.2 x (5 - 0)= mi x 4.2 x 5= mi x 21From the principle of calorimetry, if the system is fully insulated then,Heat gained by cold body = Heat lost by hot body.Q1 + Q2 = Q3 + Q4Substituting the values we get,17010+540=(mi×330)+(mi×21)17550=mi×351⇒mi=17550351⇒mi=50g17010 + 540 = (m_i \times 330) + (m_i \times 21) \[0.5em] 17550 = m_i \times 351 \[0.5em] \Rightarrow m_i = \dfrac{17550 }{351} \[0.5em] \Rightarrow m_i = 50 g17010+540=(mi​×330)+(mi​×21)17550=mi​×351⇒mi​=35117550​⇒mi​=50gHence, mass of ice = 50 g

Question 15

Question 11250 g of water at 30° C is contained in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and it's contents to 5° C. Given specific latent heat of fusion of ice = 336 x 103 J kg-1, specific heat capacity of copper = 400 J kg-1 K-1, specific heat capacity of water is 4200 J kg-1 K-1.

Accepted Answer

Given,mcopper = 50 gmwater = 250 gFinal temperature = 5° C.Let mass of ice required be mi.Heat energy gained by ice at 0° C to convert into water at 0° C= mi × L= mi x 336 JHeat energy gained by (m) g of water at 0° C to rise it's temperature to 5° C= m x c x change in temperature= mi x c x (5 - 0) = mi × 4.2 × 5= 21 x miHeat energy lost by water at 30° C in cooling to 5° C= m x c x change in temperature= 250 × 4.2 × (30 - 5)= 250 × 4.2 × 25= 26250 JHeat energy lost by vessel at 30° C to cool down to 5° C == m x c x change in temperature= 50 × 0.4 × (30 - 5)= 50 × 0.4 × 25= 500 JIf there is no loss of heat,Heat energy gained = Heat energy lostSubstituting the values in the relation above we get,(336×mi)+(21×mi)=26250+500357×mi=26750⇒mi=26750357⇒mi=74.93g(336 \times m_i ) + (21 \times m_i) = 26250 + 500 \[0.5em] 357 \times m_i = 26750 \[0.5em] \Rightarrow m_i = \dfrac{26750}{357} \[0.5em] \Rightarrow m_i = 74.93 g(336×mi​)+(21×mi​)=26250+500357×mi​=26750⇒mi​=35726750​⇒mi​=74.93gHence, required mass of ice = 74.93 g

Question 16

Question 15200 g of ice at 0° C converts into water at 0° C in 1 minute when heat is supplied to it at a constant rate. In how much time, 200 g of water at 0° C will change to 20° C? Take specific latent heat of ice = 336 J g-1.

Accepted Answer

Given,mi = 200 gmw = 200 gTime for ice to melt (t1) = 1 min = 60 sChange in temperature of water (Δt) = 20° Cspecific latent heat of ice = 336 J g-1Rate of heat exchange is constant.Therefore, power required for converting ice to water = power required to increase the temperature of water.Pi=PwEit1=Ewt2mi×Lt1=mw×cw×Δtt2200×33660=200×4.2×20t2⇒t2=5040336⇒t2=15sP_i = P_w \[0.5em] \dfrac{E_i}{t_1} = \dfrac{E_w}{t_2} \[0.5em] \dfrac{ m_i \times L}{t_1} = \dfrac{ m_w \times c_w \times Δt }{t_2} \[0.5em] \dfrac{ 200 \times 336}{60} = \dfrac{ 200 \times 4.2 \times 20 }{t_2} \[0.5em] \Rightarrow t_2 = \dfrac{5040}{336} \[0.5em] \Rightarrow t_2 = 15 sPi​=Pw​t1​Ei​​=t2​Ew​​t1​mi​×L​=t2​mw​×cw​×Δt​60200×336​=t2​200×4.2×20​⇒t2​=3365040​⇒t2​=15sHence, time taken = 15 s

Question 17

Question 16During exercise, the body loses heat through evaporation of sweat. If a person loses 1 kg of sweat during exercise, how much energy does the body lose through evaporation? How does the cooling effect of evaporation compare to heat loss due to specific heat capacity? (Latent heat of vaporization = 2268 × 10³ J kg⁻¹, Specific heat capacity of water = 4.2 × 10³ J kg⁻¹ K⁻¹)

Accepted Answer

Given,Mass of sweat lost (m) = 1 kgLatent heat of vaporization (L) = 2268 × 10³ J kg⁻¹Specific heat capacity of water (c) = 4.2 × 10³ J kg⁻¹ K⁻¹Now,Energy lost by body through evaporation (Q) = mL = 1×2268×103=2268×103 J1	imes2268	imes10^3=2268	imes10^3\ J1×2268×103=2268×103 JAnd,To raise 1 kg of water by 1 °C, energy required (q) = mcΔtOn putting values,q=1×4.2×103×1=4.2×103 Jq=1	imes4.2	imes10^3	imes1= 4.2	imes10^3\ Jq=1×4.2×103×1=4.2×103 JSo, evaporation of 1 kg of sweat causes heat loss equivalent to increasing the temperature of 1 kg of water by :Equivalent temperature rise = Qq=2268×1034.2×103=540 oC\dfrac{Q}{q}=\dfrac{2268	imes 10^3}{4.2	imes 10^3}=540\ ^oCqQ​=4.2×1032268×103​=540 oCThe cooling effect of evaporation of 1 kg of sweat is equivalent to cooling 1 kg of water by 540°C. This shows that evaporation causes a much greater cooling effect compared to simple heating or cooling based on specific heat capacity.

Question 18

Question 17The temperature of surroundings starts falling when ice in a frozen lake starts melting. Give reasons.

Accepted Answer

The temperature of surroundings starts falling when ice in a frozen lake starts melting because quite a large amount of heat energy is required for melting the frozen lake which is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it becomes very cold.

Question 19

Question 18Water in lakes and ponds do not freeze at once in cold countries. Give reason.

Accepted Answer

Water in lakes and ponds do not freeze at once in cold countries because the specific latent heat of fusion of ice is sufficiently high (= 336 J g-1). The water in lakes and ponds will have to liberate a large quantity of heat to the surrounding before freezing. The layer of ice formed over the water surface, being a poor conductor of heat, will also prevent the loss of heat from the water of lake, hence the water does not freeze all at once.

Calorimetry

Exercise 11A Numericals

Exercise 11A Short Questions

Exercise 11A Very Short Questions

Exercise 11B Numericals

Exercise 11B Short Questions