Question

What mass of steam at 100°C must be mixed with 119 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 57.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg?

Answer 1

Answer:$M=27.92\ gm$

Explanation:

Given

mass of ice $m=119\ gm$

Final temperature of liquid $T_f=57^{\circ}C$

Specific heat of water $c=4186\ J/kg-K$

Latent heat of fusion $L=333\ kJ/kg$

Latent heat of vaporization $L_v=2256\ kJ/kg$

Suppose M is the mass of steam at $100^{\circ} C$

Heat required to melt ice and convert it to water at $57^{\circ}C$

$Q_1=mL+mc(T_f-0)$

Heat released by steam

$Q_2=ML_v+Mc(100-T_f)$

$Q_1$ and $Q_2$ must be equal as the heat gained by ice is equal to Heat released by steam

$Q_1=Q_2$

$\Rightarrow mL+mc(T_f-0)= ML_v+Mc(100-T_f)$

$\Rightarrow M=\dfrac{m[L+c\times T_f]}{L_v+c(100-T_f)}$

$\Rightarrow M=\dfrac{119[333\times 10^3+4186\times 57]}{2256\times 10^3+4186\times (100-57)}$

$\Rightarrow M=119\times 0.2346$

$M=27.92\ gm$