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

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

Step-by-step 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