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

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.