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

Answer:

$Q=977216.256\ J=977.216\ kJ$

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:

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

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

$Q=384(2256+4.186\times (100-31))$

$Q=977216.256\ J=977.216\ kJ$