Question

$1\, kg$ of ice at $-20^{\circ} C$ is mixed with $2\, kg$ of water at $90^{\circ} C$. Assuming that there is no loss of energy to the environment, what will be the final temperature of the mixture? (Assume, latent heat of ice $=334.4 \,kJ / kg$, specific heat of water and ice are $4.18 \,kJ \,kg ^{-1} K ^{-1}$ and $2.09\, kJ\, kg ^{-1}- K ^{-1}$, respectively.)

• A. $30^{\circ} C$
• B. $0^{\circ} C$
• C. $80^{\circ} C$
• D. $45^{\circ} C$

Answer 1

Answer: (A)

Let final temperature of mixture is $T^{\circ} C$. Then,
Heat lost by $2 \,kg$ water at $90^{\circ} C$ to cool down at $T^{\circ} C =$ Heat gained by $1\, kg$ ice at $-20^{\circ} C$ to reach at $0^{\circ} C +$ Heat gained by $1 \,kg$ ice at $0^{\circ} C$ to change its state from ice to water + Water $1\, kg$ formed at $0^{\circ} C$ is now absorbs heat to reach temperature of $T^{\circ} C$
$\Rightarrow m_{w} s_{w} \Delta T=m_{i} s_{i}\left(0-\left(-20^{\circ} C \right)\right)$
$\Rightarrow 2 \times 4.18 \times(90-T)=1 \times 2.09 \times 20$
$+1 \times 334.4+1 \times 4.18 \times T$
$\Rightarrow 7524-376.2=3 \times 4.18 \times T$
$T=30^{\circ} C$
So, final temperature of mixture is $30^{\circ} C$.