Question 5.10: What will be the final temperature if we add 1000. cal of he...
What will be the final temperature if we add 1000. cal of heat to 10.0 g of ice at 0°C?
Strategy
The first thing the added heat does is to melt the ice. So we must first determine if 1000 cal is sufficient to melt the ice completely. If less than 1000. cal is required to melt the ice to liquid water, then the remaining heat will serve to raise the temperature of the liquid water. The specific heat (SH; Section 1.9) of liquid water is 1.00 cal/g · °C (Table 1.4).
Table 1.4 Specific Heats for Some Common Substances
| Specific Heat (cal/g · °C) | Substance | Specific Heat (cal/g · °C) | Substance |
| 0.42 | Wood (typical) | 1 | Water |
| 0.22 | Glass (typical) | 0.48 | Ice |
| 0.2 | Rock (typical) | 0.48 | Steam |
| 0.59 | Ethanol | 0.11 | Iron |
| 0.61 | Methanol | 0.22 | Aluminum |
| 0.56 | Ether | 0.092 | Copper |
| 0.21 | Carbon tetrachloride | 0.031 | Lead |
Solution
Step 1: This phase change will use 10.0 g \times 80. cal/g = 8.0 \times 10^2 cal, which leaves 2.0 \times 10^2 cal to raise the temperature of the liquid water.
Step 2: The temperature of the liquid water is now raised by the remaining heat. The relationship between specific heat, mass, and temperature change is given by the following equation (Section 1.9):
Amount of heat = SH \times m \times (T_2 – T_1)
Solving this equation for T_2 – T_1 gives:
T_2 – T_1 = amount of heat \times \frac{1}{SH} \times \frac{1}{m}
T_2 – T_1=2.0 \times 10^2 \cancel{cal} \times \frac{\cancel{g} . ^\circ }{1.00 \cancel{cal}} \times \frac{1}{10.0 \cancel{g}} =20.°C
Thus, the temperature of the liquid water will rise by 20°C from 0°C, and it is now 20°C.