Question

Calculate the quantity of heat required to convert $10kg$ of ice at $0⁰C$ to water at $50⁰C$: (Specific heat of water is $4200 J/Kg⁰C$, Latent heat of fusion of ice is $336 \times 10^3 J/Kg$).
(A) 4506 kJ
(B) 3360 kJ
(C) 5460 kJ
(D) 5400kJ

Answer 1

Hint
According to the question there is $10kg$ of ice. We have to convert the ice to $50⁰C$ of water, as we know the latent heat is the heat required to convert a solid into a liquid or vapour, or a liquid into a vapour, without change of temperature.

Complete step by step answer
We know that heat is required to change the state of a matter. Heat required to change a solid into liquid, we know that,
$Q = mL$ … (1)
where $L$= latent heat of fusion,
Now given that, $m=10 kg$ [ here $m$ = mass of ice],
and latent heat of fusion of ice is, $L=336 \times 10³J/Kg$,
hence, heat required to convert the $10kg$ of ice at $0⁰C$ to water at $0⁰C$,
now, heat required to convert the $10kg$ of water at $0⁰C$ to water at $50⁰C$, by,
$Q_{\prime} = m \times c \times \Delta t$,
given $c=4200 J/kg⁰C$,
hence $Q\prime = 10 \times 4200 \times \left( {50 - 0} \right) = 2100{\text{ }}kJ$
therefore total heat required to convert $10kg$ of ice at $0⁰C$ to water at $50⁰C$
$Q_{\prime \prime} = Q + Q_{\prime} = 3360 + 2100 = 5460{\text{ }}kJ$.
Option (C) is correct.

Note
Specific heat, ratio of the quantity of heat required to raise the temperature of a body one degree to that required to raise the temperature of an equal mass of water one degree-
$Q\prime = m \times c \times \Delta t$
Latent heat of fusion, also known as enthalpy of fusion.