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.

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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

Calorimetry — Question 15

Question 15