Question

What is the specific heat capacity of a 250, g,

250g object that needs 34, point, 125, k, J,

34.125kJ to have its temperature raised by 65, point, 0, degrees, C,

65.0∘

C?

Answer 1

The specific heat capacity is given as follows:

$$c = \dfrac{Q}{m\Delta T}$$

where $Q = 34.125\times 10^{3}J$ is the amount of heat given to the object, $\Delta T = 65\degree C$ is the temperature change, and $m = 250g = 0.25kg$ is the mass of the object. Thus, obtain:

$$c = \dfrac{34.125\times 10^3 J}{0.25kg\cdot 65\degree C} = 2.1\times 10^3\dfrac{J}{kg\cdot \degree C}$$

Answer. $2.1\times 10^3\dfrac{J}{kg\cdot \degree C}$.