Question: At a party, 4.00 kg of ice at -4.00°C is added to a cooler holding...
Question:
At a party, 4.00 kg of ice at -4.00°C is added to a cooler holding 30 liters of water at 20.0°C. What is the temperature of the water when it comes to equilibrium?
Here's a worked out solution to something similar to help you:
SOLUTION
Calculate the mass of liquid water.
mwater = ρwaterV
= (1.00 
103
kg/m3)(30.0 L)(1.00 m3 / 1.00 103
L) = 30.0 kg
Write the equation of thermal equilibrium.
(1)
Qice + Qmelt + Qice-water + Qwater = 0Construct a comprehensive table.
| Q | m (kg) | c (J/kg · °C) | L (J/kg) | Tf | Ti | Expression |
| Qice | 6.00 | 2090 | - | 0 | −5.00 | micecice(Tf − Ti) |
| Qmelt | 6.00 | 3.33
105 |
0 | 0 | miceLf | |
| Qice-water | 6.00 | 4190 | - | T | 0 | micecwat(Tf − Ti) |
| Qwater | 30.0 | 4190 | - | T | 20.0 | mwatcwat(Tf − Ti) |
Substitute all quantities in the second through sixth columns into the last column and sum, which is the evaluation of Equation (1), and solve for T.
6.27 104 J
+ 2.00 106
J
+
(2.51 104
J/°C)(T − 0°C)
+
(1.26 105
J/°C)(T − 20.0°C) = 0
T = 3.03°C
Homework Answers
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Question:
At a party, 4.00 kg of ice at -4.00°C is added to a cooler
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