Question

How much heat is required to convert solid ice with a mass of 760 g at a temperature of -16.0 °C to liquid water at a temperature of 76.5 °C? The specific heat of ice is c_ice = 2100 J/kgK, the specific heat of water is c_water = 4186.8 J/kgK, and the heat of fusion for water is L_f = 334 kJ/kg.

Answer 1

There are three steps.

(i) Temperature from -16°C to 0°C

(ii) Melting the ice.

(iii)Temperature from 0°C to 76.5°C

Given m = 760g = 0.760kg

T1 = -16°C = -16 +  273.15 = 257.15 k

T2 = 0°C = 273.15K

T3 = 76.5°C = 76.5 + 273.15 = 349.65

H1 = m x(T2 - T1) x Cice = 0.760 x {0-257.15} x 2100 = -410411.4 J

H2 = m x Lf = 0.760 x 334 = 253.84 J

H3 = m x (T3 - T2) x Cwater = 0.760 x (349.65-0) x 4186.8 = 1112575.1 J

So, total heat

H = H1 + H2 + H3 = -410411.4 + 253.84 + 1112575.1 = 702417.95 J

Hope this will help you. Have a good day. Thanks.