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Chapter 7: Problem 2

Calculate the final temperature that results when (a) a 12.6 g sample of water at \(22.9^{\circ} \mathrm{C}\) absorbs \(875 \mathrm{J}\) of heat; (b) a 1.59 kg sample of platinum at \(78.2^{\circ} \mathrm{C}\) gives off \(1.05 \mathrm{kcal}\) of heat \(\left(\mathrm{sp} \mathrm{ht} \text { of } \mathrm{Pt}=0.032 \mathrm{cal} \mathrm{g}^{-1}\right.\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\).

Short Answer

Expert verified
The final temperature of the water after it absorbs 875 J of heat is calculated to be approximately 37.8°C. The final temperature of the platinum after it gives off 1.05 kcal of heat is calculated to be approximately 76.3°C.

Step by step solution

01

Calculating Final Temperature of Heated Water

To calculate the final temperature after heating the water, the heat capacity equation is used. It is given that the water absorbs 875 J of heat, the mass is 12.6 g, the initial temperature is 22.9°C, and the specific heat of water is 4.18 J/g°C. The equation will be set up as \(875 = 12.6 \times 4.18 \times (T_f - 22.9)\), where \(T_f\) is the final temperature. Solving for \(T_f\) will give the final temperature of the water after heating.
02

Calculating Final Temperature of Cooled Platinum

To calculate the final temperature after cooling the platinum, the heat capacity equation is used again. It is given that the platinum gives off 1.05 kcal of heat (which needs to be converted to joules for consistent units), the mass is 1.59 kg (which needs to be converted to grams for consistent units), the initial temperature is 78.2°C, and the specific heat of platinum is 0.032 cal/g°C (which needs to be converted to J/g°C for consistent units). The equation will be set up as \(-(1.05 \times 4184) = (1590 \times 0.032 \times 4.18) \times (T_f - 78.2)\), where \(T_f\) is the final temperature. Solving for \(T_f\) will give the final temperature of the platinum after cooling.

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