Question

What mass of steam at 100°C must be mixed with 499 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 33.0°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 the steam is 91.2 g

Mass of the steam=91 grams

Step-by-step explanation:

Given:

• Mass of the ice=499 g
• Final temperature of the liquid water
  `=33^\circ \rm C`
• Latent heat of fusion=
  `333\ \rm kJ\kg`

• Latent heat of vaporization =
  `2256\ \rm kJ\kg`

When steam is mixed with the ice then the heat loss by the steam will be gained by the ice so there will no overall heat gain or loss during the mixing

So According to question

Let M be the mass of the steam mixed with ice then we have

`M*2256*10^3+M*4186*(100-33)=0.499*222*10^3+0.499*4186*(33-0)\\M*2.58*10^6=2.35*10^5\\M=91.2\ \rm g`