Question

Using the values for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following:
A. Joules released when 125g of steam at 100 C condenses and cools to liquid at 15.0 C.
B. Kilocalories needed to melt a 525-g ice sculpture at 0 C and to warm the liquid to 15.0 C
C. Kilojoules released when 85.0 g of steam condenses at 100 C, cools, and freezes at 0 C

Answer 1

A. Joules released when 125g of steam at 100 C condenses and cools to liquid at 15.0 C.

We require 2 type of heat, latent heat and sensible heat

Sensible heat (CP): heat change due to Temperature difference

Latent heat (LH): Heat involved in changing phases (no change of T)

Then

Q1 = m*Cp ice * (Tf – T1)

Q2 = m*LH ice

Q3 = m*Cp wáter * (Tb – Tf)

Q4 = m*LH vap

Q5 = m*Cp vap* (T2 – Tb)

Note that Tf = 0°C; Tb = 100°C, LH ice = 334 kJ/kg; LH water = 2264.76 kJ/kg.

Cp ice = 2.01 J/g°C ; Cp water = 4.184 J/g°C; Cp vapor = 2.030 kJ/kg°C

Then

Q1 = m*2.01 * (0 – T1) --> no ice forms

Q2 = m*334 --> no freezing involved

Q3 = m*4.184 * (100 – 15)

Q4 = m*2264.76

Q5 = m*2.03* (T2 – 100) --> T2 = 100°C so no sensible hat of vapor

we need:

Q3 = m*4.184 * (100 – 15) = 125 * 4.184 * (15-100) = -44455

Q4 = m*2264.76 = -125 *2264.76 = -283095

Qt = Q3+Q4 = 44455 +283095 = releaed 327550 J

b)

we need:

Q2 = m*LH ice

Q3 = m*Cp wáter * (Tb – Tf)

substitute

Q2 = 525*334 = 175350

Q3 = 525*4.184* (15– 0) = 32949

Qtotal = 175350 +32949 = 208299 J = 208.299 kJ

C)

we need:

Q2 = m*LH ice

Q3 = m*Cp wáter * (Tb – Tf)

Q4 = m*LH vap

substitute

Q2 = 85*334 = 28390

Q3 = 85*4.184 * (100-0) = 35564

Q4 = 85*2264.76 = 192504.6

Qtotal= 28390 + 35564 + 192504.6

Qtotal = 256458.6 J

Q in kJ = 256458.6 /1000 kJ = 256.45 kJ