Question

A $5g$ piece of ice at $-{20}^{o}C$ is added into $50g$ of water at ${20}^{o}C$. Find the temperature of the mixture.($C_{ice}=0.5cal/g{m}^{o}C$ and $L=80J/gm$)

• A. ${10}^{o}C$
• B. ${12.6}^{o}C$
• C. ${13.1}^{o}C$
• D. ${14.2}^{o}C$

Answer 1

The correct option is A ${10}^{o}C$

Let the temperature of mixture be $\theta$. Heat lost by hot water $=$ heat gained by ice.
$m_W{C}_W(20-\theta )=m{C}_{ice}\Delta \theta +mL+m{C}_W\theta$
$50\times 1(20-\theta )=5\left(20\right)\times 0.5+5\times 80+5\times 1\times \theta$
$55\theta =1000-450=550$
$\theta ={10}^{o}C$