Question

A piece of ice (heat capacity$=2100J/k{g}^{\circ }C$ and latent heat $=3.36\times {10}^{5}J/kg$) of mass $m$ grams is at $-5^{\circ }C$ at atmospheric pressure. It is given $420J$ of heat so that the ice starts melting. Finally, when the ice-water mixture is in equilibrium, it is found that $1g$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is

Answer 1

We know that: $Q=mL,Q=ms\Delta \theta$

Given: $\theta =-5^{\circ }C,Q=420J,s=2100J/k{g}^{\circ }C,L=3.36\times {10}^{5}J/kg$

Heat energy required for ice to reach $0^{\circ }C$ is

$\Delta Q_1=ms(0-(-5\left)\right)=m\times 2100\times 5\dots \left(i\right)$

Heat energy required to convert $1g$ of ice to water

$\Delta Q_2=mL=0.001\times 3.36\times {10}^5=336J$

$\Delta Q_1+\Delta Q_2=420J$

$\Delta Q_1=420-336=84J\dots \left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right)$,

$m=\frac{84}{2100\times 5}\times 1000$

$m=\frac{168}{21}=\frac{24}{3}=8g$

Final answer: $8g$.