Question

A piece of food is placed into a bomb calorimeter. How much energy (in food calories/Cal/kcal) was stored in the food if 500. g of water increases from 25 to 75°C

Answer 1

In a bomb calorimeter, the energy released from the food equals to the energy absorbed by water:

$E_{food} = Q_{water}$ .

The quantity of heat absorbed by water equals the product of the specific heat capacity of water $c$ , the mass of water $m$ (500 g) and the difference of temperature of the water $∆T$ (75°C - 25°C). Note that as we use the difference temperature, there is no need to convert from celsius to kelvin. Assuming no dependence of the specific heat of water on temperature:

$Q_{water} = c·m·∆T$

$Q_{water} = 4179.6 (\text{J/(kgK)})·500·10^{-3}(\text{kg})·(75-25)(\text{K})$

$Q_{water} = 104.5$ kJ.

Therefore, 104.5 kJ was stored in the food. Let's convert kJ to food calories/Cal/kcal:

food calories, or Cal, or kcal: $E_{fc} = E_{kJ}·0.239 = 25.0$ food calories

Answer: 25.0 food calories/Cal/kcal was stored in the food if 500. g of water increases from 25 to 75°C.