The energy content of a certain food is to be determined in a bomb calorimeter that contains 3 kg of water by burning a 2-g sample of it in the presence of 100 g of air in the reaction chamber. If the water temperature rises by 3.2°C when equilibrium is established, determine the energy content of the food, in kJ/kg, by neglecting the thermal energy stored in the reaction chamber and the energy supplied by the mixer. By calculating the rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber, do you think it is reasonable to disregard the change in the sensible energy content of the reaction chamber in the analysis? The specific heat of water at room temperature is c = 4.18 kJ/kg·°C. The constant-volume specific heat of air at room temperature is cv = 0.718 kJ/kg·°C.

Question

Moshe Middleton

Physics

27 April, 20:30

The energy content of a certain food is to be determined in a bomb calorimeter that contains 3 kg of water by burning a 2-g sample of it in the presence of 100 g of air in the reaction chamber. If the water temperature rises by 3.2°C when equilibrium is established, determine the energy content of the food, in kJ/kg, by neglecting the thermal energy stored in the reaction chamber and the energy supplied by the mixer. By calculating the rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber, do you think it is reasonable to disregard the change in the sensible energy content of the reaction chamber in the analysis? The specific heat of water at room temperature is c = 4.18 kJ/kg·°C. The constant-volume specific heat of air at room temperature is cv = 0.718 kJ/kg·°C.

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Answers (1)

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Savannah Coffey · Answer 1

20.179 x 10⁶ J / kg

Explanation:

The food after the reaction gives out heat which increases the temperature of water and air in the reaction chamber. The heat absorbed by water and air gives the estimate of energy content of the food.

Heat absorbed by water = mass x specific heat x rise in temperature

= 3 x 4.18 x 10³ x 3.2

= 40.128 x 10³ J

Heat absorbed by air = mass x specific heat x rise in temperature

0.1 x 3. 2 x. 718 x 10³

= 0.23 x 10³

Total heat energy evolved

= 40.358 x 10³ J

This energy is evolved by 2 x 10⁻³ kg of food

energy content per kg of food

= 40.358 x 10³ / 2 x 10⁻³

= 20.179 x 10⁶ J / kg