Question

A .250 kg block of pure material is heated from 20.0 degrees C to 66.0 degrees C by the addition of 5.35 kJ of energy. Calculate its specific heat, in kilocalories per kilogram per degree Celsius

Answer 1

Given:

The mass of the block of pure metal is: m = 250 kg

The change in the temperature is: ΔT = 66.0 °C - 20.0 °C = 46 °C

The heat supplied to the material is: Q = 5.35 kJ

To find:

The specific heat.

Explanation:

The expression the heat supplied/released Q, the mass of the material m, the specific heat C, and the change in temperature ΔT is given as:

$Q=mC\Delta T$

Rearranging the above equation, we get:

$C=\frac{Q}{m\Delta T}$

The heat can be converted from Kilojoules to kilocalories as:

$Q=5.35\text{ kJ}=5350\text{ J}=5350\text{ J}\times\frac{1\text{ Cal}}{4.184\text{ J}}=1278.68\text{ Cal}=1.27868\text{ KCal}$

Substituting the values in the above equation, we get:

$\begin{gathered} C=\frac{1.27868\text{ KCal}}{250\text{ kg}\times46\text{ }\degree\text{C}} \\ \\ C=1.1119\times10^{-4}\text{ KCal/kg}\degree\text{C} \end{gathered}$

Final answer:

The value of specific heat is 1.1119 × 10⁻⁴ KCal/kg °C.