Question

Using heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following.

(a) calories to condense 141 g of steam at 100

Answer 1

(a)

Step 1 steam @ 100 *C --> water @ 100 *C dH₁
Step 2 water @ 100 *C --> water @ 15 *C dH₂

Latent heat of vaporization of water is 2260 J/g

dH₁ = mLf = (141 g) x (-2260 J/g) = -3.18 x 10⁵ J

This means that 3.18 x 10⁵ J of heat was removed from the steam at 100 *C to produce water 141 g of water at 100 *C.

dH₂ = mcdT

Specific heat of water = 4.184 J/g*C
dH₂ = (141 g)(4.184 J/g*C)(15 *C - 100 *C) = -5.01 x 10^4 J

Total heat = -3.18 x 10^5 J + -5.01 x 10^4 J = -3.68 x 10^5 J

(b)

Step 1 to melt the ice @ 100*C
Step 2 to heat the water @ 15*C

dH₁ = mLf

.Latent heat of fusion (melting) of ice at 0