1 Chapter Basics of Heat Transfer
1-132 A 0.2-L glass of water at 20C is to be cooled with ice to 5C. The amounts of ice or cold water that
needs to be added to the water are to be determined.
Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water is
negligible.
Properties The density of water is 1 kg/L, and the specific heat of water at room temperature is C = 4.18
kJ/kgC (Table A-9). The heat of fusion of ice at atmospheric pressure is 333.7 kJ/kg,.
Analysis The mass of the water is
Ice,
0C
m w = V = (1kg/L )(0.2 L ) = 0.2kg
We take the ice and the water as our system, and disregard any heat and mass
transfer. This is a reasonable assumption since the time period of the process
is very short. Then the energy balance can be written as
E in E out
1 24
43
=
Net energy transfer
by heat, work, and system
1 mass
E 24
43
[mC(0 C T )
o
0 = U (U )ice + (U )water = 0
Change in internal, kinetic,
potential, etc. energies
1 solid
(
+ mhif + mC T2 0 o C
)]
liquid ice
+ [mC (T2 T1 )]water = 0
Noting that T1, ice = 0C and T2 = 5C and substituting
m[0 + 333.7 kJ/kg + (4.18 kJ/kgC)(5-0)C] + (0.2 kg)(4.18 kJ/kgC)(5-20)C = 0
It gives
m = 0.0354 kg = 35.4 g
Cooling with cold water can be handled the same way. All we need to do is
replace the terms for ice by the ones for cold water at 0C:
(U )coldwater + (U )water = 0
[mC (T2 T1 )]coldwater + [mC (T2 T1 )]water = 0
Substituting,
[mcold water (4.18 kJ/kgC)(5 - 0)C] + (0.2 kg)(4.18 kJ/kgC)(5-20)C = 0
It gives
m = 0.6 kg = 600 g
Discussion Note that this is 17 times the amount of ice needed, and it explains why we
use ice instead of water to cool drinks.
1-70
Water
0.2 L
20C