Answer to Question #114900 in Molecular Physics | Thermodynamics for Oboira Sunday

Question

0.5kg of water at 10 degree Celsius is completely converted to ice at 0 degree Celsius by extracting 188000 joules of heat from it. If the specific heat capacity of water is 4200 joules per kilogram per Kelvin, calculate the specific latent heat of fusion of ice

Answer 1

The total amount of heat taken from the water consists of two parts:

$Q_{total} = Q_{cool} + Q_{fus}$ ,

where $Q_{cool}$ is the heat that should be taken to cool the water from $10\degree C$ to $0\degree C$ and $Q_{fus}$ is the heat that should be taken to freez all water.

$Q_{cool} = c\cdot m\cdot(T_1 - T_2) = 4200*0.5*(10-0) = 21000J$.

Thus,

$Q_{fus} = Q_{total} - Q_{cool} = 188000 - 21000 = 167000.$

By definition, the specific latent heat of fusion is:

$L = \dfrac{Q_{fus}}{m} = \dfrac{167000J}{0.5kg} = 334 \dfrac{kJ}{kg}$ .

Answer. The specific latent heat of fusion is 334 kJ/kg.