Compute the quantity of heat released by 25.0 g of steam initially at 100.0oC, when it is cooled to 34.0°C and by 25.0 g of water initially at 100.0 oC, when it is cooled to 34.0°C.

For steam, heat released E = 15.26KJ

For water, heat released E2 = 6.91KJ

Explanation:

Given;

Mass (steam) ms = 25g

Mass (water) mw = 25g

Change in temperature of both steam and water ∆T = 100-34 = 66°C

Specific heat of water C = 4.186 J/g.°C

Specific Latent heat L = 334J/g

For steam;

Heat released E = msL + msC∆T

E = (25*334) + (25*4.186*66)

E = 15256.9J

E = 15.26KJ

For water;

Heat released E2 = mwC∆T

E2 = 25*4.186*66

E2 = 6906.9J

E2 = 6.91KJ