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?

# 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  of water at 25 °C?

1. A

61.6 °C

2. B

- 243.19 J

3. C

234.19 J

4. D

- 234.19 J

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### Solution:

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

Hence

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

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