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 mol1
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

Answer on Question #80953, Chemistry/ Organic Chemistry

In the calibration of a calorimeter, an electrical resistance heater supplies 80.0J of heat and a temperature increase of $1.20^{\circ}\mathrm{C}$ is observed. What is the initial temperature of $8.50\mathrm{g}$ 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 $28.0^{\circ}\mathrm{C}$ and the final temperature of the system is $50.0^{\circ}\mathrm{C}$.

Solution

Find $c_{\mathrm{cal}}$

$$q _ {c a l} = c _ {c a l} \times \Delta T, \text{ then } c _ {c a l} = \frac {q _ {c a l}}{\Delta T}$$ $$c _ {c a l} = \frac {8 0 J}{(2 7 3 . 1 5 + 1 . 2 0) K} = 0. 2 9 2 \frac {J}{K}$$

Heat absorbed by gasoline is:

$$q _ {1} = n c _ {p} \Delta T$$ $$n \left(C _ {8} H _ {1 8}\right) = m / M = 8. 5 0 g / 1 1 4 g / m o l = 0. 0 7 5 6 m o l$$ $$T _ {\text {f i n a l}} = 2 7 3, 1 5 + 5 0 = 3 2 3. 1 5 \mathrm {K}$$ $$q _ {1} = 0. 0 7 5 6 \mathrm {m o l} \times 3 9. 3 6 \mathrm {J} / (\mathrm {m o l} \times \mathrm {K}) \times (3 2 3. 1 5 - T _ {1})$$

Heat absorbed by calorimeter:

$$q _ {2} = c _ {\text {c a l}} \times \Delta T$$ $$T _ {1} = 2 8 + 2 7 3. 1 5 = 3 0 1. 1 5 \mathrm {K}$$ $$T _ {2} = 2 7 3, 1 5 + 5 0 = 3 2 3. 1 5 \mathrm {K}$$ $$q _ {2} = 0. 2 9 2 \mathrm {J} / \mathrm {K} \times (3 2 3. 1 5 - 3 0 1. 1 5) \mathrm {K} = 6. 4 2 4 \mathrm {J}$$

$q$, supplied by electrical resistance heater, is 80.0J

$q$, absorbed by gasoline and calorimeter is $q_{1} + q_{2}$

$$0. 0 7 5 6 \mathrm {m o l} \times 3 9. 3 6 \mathrm {J} / (\mathrm {m o l} \times \mathrm {K}) \times (3 2 3. 1 5 - T _ {1}) + 6. 4 2 4 \mathrm {J} = 8 0 \mathrm {J}$$ $$0. 0 7 5 6 \mathrm {m o l} \times 3 9. 3 6 \mathrm {J} / (\mathrm {m o l} \times \mathrm {K}) \times (3 2 3. 1 5 - T _ {1}) = 7 3. 5 7 6 \mathrm {J}$$ $$3 2 3. 1 5 - T _ {1} = 2 4. 7 3$$ $$T _ {1} = 2 9 8. 4 2$$ $$T \left(^ {0} C\right) = 2 9 8. 4 2 - 2 7 3. 1 5 = 2 5. 2 7 ^ {\circ} \mathrm {C}$$

Answer: $25.27^{\circ} \mathrm{C}$