Question

suppose 384g of steam originally at 100C is quickly cooled to produce liquid water at 31C. How much heat must be removed from the steam to accomplish this?

Answer 1

`Q=977216.256\ J=977.216\ kJ`

Step-by-step explanation:

Given:

• mass of steam,
  `m=384\ g`

• temperature of steam,
  `T_(is)=100^(\circ)C`

• temperature of resultant water,
  `T_(fw)=31^(\circ)C`

We have,

• latent heat of vapourization of water,
  `L=2256\ J.g^(-1)`

• specific heat capacity of water,
  `c=4.186\ J.g^(-1)`

When we cool the steam of 100°C then firstly it loses its latent heat to convert into water of 100°C and the further cools the water.

Now the heat removed from steam to achieve the final state of water:

`\rm Q=latent\ heat\ of\ vapourization+sensible\ heat\ of\ water`

`Q=m(L+c.\Delta T)`

`Q=384(2256+4.186* (100-31))`

`Q=977216.256\ J=977.216\ kJ`