Question

A calorimeter of mass 60g contains 180g of water at 29°C . Find the final temperature of the mixture when 37.2g of ice at -10°C is added to it( specific heat capacity of water=4200J/kg/k, latent heat of ice=336×10^3 J/kg, specific heat capacity of ice=2100J/kg/k, specific heat capacity of the calorimeter is 0.42J/g/°C

Answer 1

Let final temperature =T

Change in temperature of calorimeter and water = 29°C - T since the final temperature is lower than the initial temperature because temperature change involves losing heat.

Change in temperature of the ice= T +10°C since it involves gaining of heat

Heat gained by ice= Heat lost by water + heat lost by calorimeter.

0.0372 × 336×10^3 + 0.0372× 2100 ×(T + 10) =0.18 × 4200 ×(29-T) + 0.42 × 0.060 × (29-T)

12499.2 + 78.12T + 781.2= (21924-756T) + (0.7308-0.0252T)

-8644.3308 = - 834.1452T

T=10.363°

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