Question

$100$ g of ice is mixed with $100$ g of water at ${100}^{\circ }$C. What will be the final temperature of the mixture?

• A. ${13.33}^{\circ }C$
• B. ${20}^{\circ }C$
• C. ${23}^{\circ }C$
• D. ${40}^{\circ }C$

Answer 1

Answer: (A)

${13.33}^{\circ }C$

Let the final temperature of the mixture be T.

Then $100\times 80+100(T-0)\times 1/2100\times 80+100(T-0)\times 1/2$

(as specific heat of ice is 0.5cal/g/℃ and specific heat of water is 1cal/g∘C)

$=100\times 1\times (100-T)$

Solving, we get $T=13.33℃$