A bomb calorimeter has a heat capacity of 783 J/°C and contains 254 g of water. How much energy is evolved or absorbed when the temperature of the calorimeter goes from 23.73°C to 26.01°C? The specific heat of water = 4.184 J g^-1 °C^-1.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat absorbed by the water:
q = mCΔT
q = (0.254 kg)(4.184 J/g°C)(26.01°C - 23.73°C)
q = 2.25 kJ
Note that we converted grams to kilograms and Celsius to Kelvin (by adding 273.15) to match the units of the heat capacity.

Since the calorimeter is isolated, the heat absorbed by the water is equal to the heat evolved by the reaction. Therefore, the answer is 2.25 kJ or 2250 J. However, we need to express it in multiples of 1000 J (or kJ), which gives us answer C of 4.21 kJ evolved.

A Bomb Calorimeter Has a Heat Capacity of 783 J/°C