If one gram of kerosene liberates 46.0 kJ of heat when it is burned, to what temperature can 0.25 g of kerosene raise the temperature of 75 cm3 of water at 25 °C?

A
61.6 °C
B
- 243.19 J
C
234.19 J
D
- 234.19 J

Solution:

Heat liberated $= \frac{1}{4} (46.0 kJ) = 11.5 kJ = (11500 / 4.184) cal$

Hence

$(75 g) (1 {cal}^{\circ} C^{- 1} g^{- 1}) (t - 25^{\circ} C) = \frac{11500}{4.184} cal$ i.e. $t - 25^{\circ} C = {\frac{11500}{4.184 \times 75}}^{\circ} C = {36.65}^{\circ} C$

$t = {36.65}^{\circ} C + 25^{\circ} C = {61.65}^{\circ} C$

If one gram of kerosene liberates 46.0 kJ of heat when it is burned, to what temperature can 0.25 g of kerosene raise the temperature of $75 {cm}^{3}$ of water at 25 °C?

Solution:

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