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`

Step-by-step 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=(m[L+c* T_f])/(L_v+c(100-T_f))`

`\Rightarrow M=(119[333* 10^3+4186* 57])/(2256* 10^3+4186* (100-57))`

`\Rightarrow M=119* 0.2346`

`M=27.92\ gm`