Question

$10g$ of ice at ${10}^{0}C$ is mixed with $100g$ of water at ${50}^{0}C.$ What is the resultant temperature of mixture

• A. ${31.2}^{0}C$
• B. ${32.8}^{0}C$
• C. ${36.7}^{0}C$
• D. ${38.2}^{0}C$

Answer 1

The correct option is D ${38.2}^{0}C$

let resultant temp. $=x$

for $10g$ water $m=10g$

$\Delta t=(10-x)$

Now

$Q=m.c.\Delta t$

$Q=\mathrm{10.1.}(10-x)$

here

when $100g$ of water

$m=100g$

$t=(50-x)$

$Q=100(50-x)$

heat gained = heat lost $10(10-x)=100(50-x)$

$x={38.2}^{\circ }C$