Question

If 10 g of ice is added to 40 g of water $150^{o}C,$ then the temperature of the mixture is (specific heat of water $= 4.2 \times 10^{3}Jkg^{- 1}K^{- 1}$,Latent heat of fusion of ice $= 3.36 \times 10^{5}Jkg^{- 1}$)

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

Answer 1

Answer: (D)

$0^{o}C$

Answer 2

Heat lost by water to come from $15{^\circ}Cto0{^\circ}C$is

$H_{1} = ms\Delta T\frac{40}{100} \times (4.2 \times 10^{3}) \times (15 - 0) = 2520J$

Heat required to convert 10 g ice into 10g water at $0{^\circ}C$is $H_{2} = mL = \frac{10}{1000} \times (3.36 \times 10^{5}) = 3360J$

Since $H_{2} > H_{1}$, 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 $0{^\circ}C.$