Question 9(c)A piece of ice of mass 60 g is dropped into 140 g of water at 50 °C. [4]Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding).Specific heat capacity of water (c) = 4.2 Jg-1k-1,Specific latent heat of fusion of ice (L) = 336 Jg-1

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Let final temperature of water be T °C.Heat energy taken by the ice to melt at 0 °CQ1 = mass of ice x specific latent heat of fusion of ice= 60 x 336 = 20,160 JHeat energy taken by the melted ice to raise it's temperature from 0 °C to T °C.Q2 = mass of ice x specific heat capacity of water x rise in temperature= 60 x 4.2 x (T - 0) = 252T JTotal heat energy taken by ice = Q1 + Q2= (20,160 + 252 T) JHeat energy given by water in fall of it's temperature from 50 °C to T °C= mass of water x specific heat capacity of water x fall in temperature= 140 x 4.2 x (50 - T)= 588 (50 - T) JIf there is no heat loss,heat energy given by water = total heat energy taken by iceSubstituting the values we get,588(50−T)=20,160+252T(588×50)−588T=20,160+252T29,400−588T=20,160+252T29,400−20,160=588T+252T9240=840T⇒T=9240840⇒T=11°C588 (50 - T) = 20,160 + 252 T \[0.5em] (588 \times 50) - 588 T = 20,160 + 252 T \[0.5em] 29,400 - 588 T = 20,160 + 252 T \[0.5em] 29,400 - 20,160 = 588 T + 252 T \[0.5em] 9240 = 840 T \[0.5em] \Rightarrow T = \dfrac{9240}{840} \[0.5em] \Rightarrow T = 11 °C588(50−T)=20,160+252T(588×50)−588T=20,160+252T29,400−588T=20,160+252T29,400−20,160=588T+252T9240=840T⇒T=8409240​⇒T=11°CHence, T = 11 °C

Solved 2020 Question Paper ICSE Class 10 Physics — Question 15

Question 9(c)