Question

150g of ice at 0°c is mixed with 300g of water at 50°c.calculate the temperature of the mixture.Take specific heat capacity of water as 4.2×10^3J/Kg/k (specific latent heat of fussion of ice as 3.34×10^5J/Kg/k

Answer 1

The energy ballance for such situation will be the following:

$$Q_{loss} = Q_{melt} + Q_{heat}$$

where $Q_{loss}$ is the energy lost by hot water, $Q_{melt}$ is the energy required to melt the given amount of ice, $Q_{heat}$ is the enrgy required to heat the cold water obtained after melting.

Writting the expression for $Q_{loss}$

$$Q_{loss} = m_{hot}\cdot c_{water}(T_0 - T)$$

where $m_{hot} = 0.3kg$, $c_{water} = 4.2\cdot 10^{3} \dfrac{J}{kg\cdot K}$, $T_0 = 50\degree C = 323K$ and $T$ is required final temperature.

Writting the expression for $Q_{melt}$:

$$Q_{melt} = m_{ice}\lambda$$

where $m_{ice} = 0.15 g$ and $\lambda = 3.34\cdot 10^5 \dfrac{J}{kg}$.

The expression for $Q_{heat}$ will be:

$$Q_{heat} = m_{ice}\cdot c_{water}(T-273 K)$$

Combining it all together:

$$m_{hot}\cdot c_{water}(T_0 - T) = m_{ice}\lambda + m_{ice}\cdot c_{water}(T-273 K)$$

Expressing $T$ from the last equation, obtain:

$$T = \dfrac{m_{hot}T_0 + m_{ice}\cdot 273K}{m_{hot} + m_{ice}} - \dfrac{m_{ice}\lambda}{c(m_{hot} + m_{ice})}$$

Substituting numerical values:

$$T = \dfrac{0.3\cdot 323 + 0.15\cdot 273}{0.3+0.15} - \dfrac{0.15\cdot 3.34\cdot 10^5}{ 4.2\cdot 10^{3}(0.3 + 0.15)} \approx 279.8K = 6.8 \degree C$$

Answer. T = 6.8°C.