Question

Four cubes of ice at -${10}^{\circ }C$ each $onegm$ is taken out from the refrigerator and are put in $150gm$ of water at ${20}^{\circ }C$. The temperature of water when thermal equilibrium is atttained:
Assume that no hate is lost to the outside and water equivalent of container is $46gm$. (Specific heat capacity of water is $1cal/g^{\circ }C$, specific heat capacity of ice is $0.5cal/g^{\circ }C$, Latent heat of fusion of ice $80cal/g^{\circ }C$)

• A. $0^{\circ }C$
• B. None
• C. $-{10}^{\circ }C$
• D. ${17.9}^{\circ }C$

Answer 1

The correct option is D ${17.9}^{\circ }C$

Formula used: $Q=mL,Q=ms\Delta T$
Given: ${\theta }_{ice}=-{10}^{\circ }C,m_{ice}=1g,m_{water}=150g,{\theta }_2={20}^{\circ }C,C_{container}=46g,s_w=1cal/g^{\circ }C,s_{ice}=0\cdot 5cal/g^{\circ }C,L_f=80cal/g$

Let final temperature of mixture be $T$.
Heat gained by ice is equal to heat lost by water + heat lost by container. (initial temperature of container at ${20}^{\circ }C$)
$m_{ice}{s}_{ice}(0-(-10))+m_{ice}{L}_f+m_{ice}{s}_w(T-0)=m_w{s}_w(20-T)$
$4\times \frac{1}{2}\times 10+4\times 80+4\times 1\times (T-0)=196\times 1\times (20-T)$
$20+320+4T=196\times 20-196T$
$200T=196\times 20-340$
$T=\frac{3580}{200}={17.9}^{\circ }C$

FINAL ANSWER: (c).