Question

What mass of steam at 100°C must be mixed with 490 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 89.0°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:

the mass of steam at 100°C must be mixed is 150 g

Explanation:

given information:

ice's mass, $m_{i}$ = 490 g = 0.49 kg

steam temperature, T = 100°C

liquid water temperature, T = 89.0°C

specific heat of water, $c_{w}$ = 4186 J/kg.K = 4.186 kJ/kg.K

latent heat of fusion, $L_{f}$ = 333 kJ/kg

latent heat of vaporization, $L_{v}$ = 2256 kJ/kg

first, we calculate the heat of melted ice to water

Q₁ = $m_{i} L_{f}$

where

Q = heat

$m_{i}$ = mass of the ice

$L_{f}$  = latent heat of fusion

thus,

Q₁ = $m_{i} L_{f}$

    = 0.49 x 333

    = 163.17 kJ

then, the heat needed to increase the temperature of water to 89.0°C

Q₂ = $m_{i}$ $c_{w}$ (89 - 0), the temperature of ice is 0°C

$c_{w}$ = specific heat of water

so,

Q₂ = $m_{i}$ $c_{w}$ (89 - 0)

     = 0.49 x 4.186 x 89

     = 182.55 kJ

so, the heat absorbed by the ice is

Q = Q₁ + Q₂

   = 163.17 + 182.55

   = 345.72 kJ

the temperature of the steam is 100°C, so the mass of the steam is

Q = $m_{s}$$L_{v}$  +  $m_{s}$$c_{w}$ (100 - 89)

Q = $m_{s}$($L_{v}$  +  $c_{w}$ (11))

$m_{s}$ = Q/ [$L_{v}$  +  $c_{w}$ (11)]

      = 345.72/ [2256 + (4.186 x 11)]

      = 0.15 kg

      = 150 g