What will be the final temperature if we add 1000. cal of heat to 10.0 g of ice at 0°C?

Question

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

Accepted Answer

Solution
Step 1: This phase change will use 10.0 g [latex] imes [/latex] 80. cal/g = 8.0 [latex] imes 10^2[/latex] cal, which leaves 2.0 [latex] imes 10^2[/latex] 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 [latex] imes m imes (T_2 - T_1) [/latex]

Solving this equation for [latex]T_2 - T_1[/latex] gives:

[latex]T_2 - T_1[/latex] = amount of heat [latex] imes \frac{1}{SH} imes \frac{1}{m} [/latex]

[latex]T_2 - T_1=2.0 imes 10^2 \cancel{cal} imes \frac{\cancel{g} . ^\circ }{1.00 \cancel{cal}} imes \frac{1}{10.0 \cancel{g}} [/latex]=20.°C

Thus, the temperature of the liquid water will rise by 20°C from 0°C, and it is now 20°C.

Question 5.10: What will be the final temperature if we add 1000. cal of he...

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