Question

In the calibration of a calorimeter, an electrical resistance heater supplies 80.0J of
heat and a temperature increase of 1.20°C is observed. What is the initial
temperature of 8.50 g of gasoline (C8H18) with molar heat capacity of 39.36 J mol-
1 K-1 after heating in the same calorimeter? The initial temperature of the
calorimeter is 28.0°C and the final temperature of the system is 50.0°C

Answer 1

Find $c_{\mathrm{cal}}$

$$q_{\mathrm{cal}} = c_{\mathrm{cal}} \cdot \Delta T, \text{ then } c_{\mathrm{cal}} = \frac{q_{\mathrm{cal}}}{\Delta T}$$ $$c_{\mathrm{cal}} = \frac{80 \, J}{(273.15 + 1.20) \, K} = 0.292 \, \frac{J}{K}$$

Heat absorbed by gasoline is:

$$q_1 = n c_p \Delta T$$ $$n(C_8H_{18}) = m / M = 8.50 \, \mathrm{g} / 114 \, \mathrm{g} / \mathrm{mol} = 0.0756 \, \mathrm{mol}$$ $$T_{\mathrm{final}} = 273.15 + 50 = 323.15 \, \mathrm{K}$$ $$q_1 = 0.0756 \, \mathrm{mol} \cdot 39.36 \, \mathrm{J} / (\mathrm{mol} \cdot \mathrm{K}) \cdot (323.15 - T_1)$$

Heat absorbed by calorimeter:

$$q_2 = c_{\mathrm{cal}} \cdot \Delta T$$ $$T_1 = 28 + 273.15 = 301.15 \, \mathrm{K}$$ $$T_2 = 273.15 + 50 = 323.15 \, \mathrm{K}$$ $$q_2 = 0.292 \, \mathrm{J} / \mathrm{K} \cdot (323.15 - 301.15) \, \mathrm{K} = 6.424 \, \mathrm{J}$$

q, supplied by electrical resistance heater, is 80.0 J

q, absorbed by gasoline and calorimeter is $q_1 + q_2$

$$0.0756 \, \mathrm{mol} \cdot 39.36 \, \mathrm{J} / (\mathrm{mol} \cdot \mathrm{K}) \cdot (323.15 - T_1) + 6.424 \, \mathrm{J} = 80 \, \mathrm{J}$$ $$0.0756 \, \mathrm{mol} \cdot 39.36 \, \mathrm{J} / (\mathrm{mol} \cdot \mathrm{K}) \cdot (323.15 - T_1) = 73.576 \, \mathrm{J}$$ $$323.15 - T_1 = 24.73$$ $$T_1 = 298.42$$ T \, (^{\circ} C) = 298.42 - 273.15 = 25.72 \, ^\circ \mathrm{C}

Answer: 25.72 \, ^\circ \mathrm{C}

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