Question

If 10 g of ice is added to $40g$ of water at $15°C$, then the temperature of the mixture is (specific heat of water = $4.2\times 10^{3}Jk{g}^{-1}{K}^{-1}$, Latent heat of fusion of ice = $3.36\times 10^{5}Jk{g}^{-1})$

• A. 15${}^{\circ }$C
• B. 12${}^{\circ }$C
• C. 10${}^{\circ }$C
• D. 0${}^{\circ }$C

Answer 1

Answer: (D)

Heat lost by water to come from 15 ${}^{\circ }$C to 0 ${}^{\circ }$C is
$H_1=ms\Delta T=\frac{40}{1000}\times \left(4.2\times 10^3\right)\times \left(15-0\right)$
= 2520 J
Heat required to convert 10 g ice into 10 g water at 0${}^{\circ }$C is
$H_2=mL=\frac{10}{1000}\times \left(3.36\times 10^5\right)=3360$ J
Since H₂ > H₁, so the whole ice will not be converted into water, whereas the temperature of the whole water will be 0${}^{\circ }$C.
Therefore the temperature of the mixture is 0${}^{\circ }$C.