When 300 J of heat is added to 25 gm of sample of a material its temperature rises from 25°C to 45°C. the thermal capacity of the sample and specific heat of the material are respectively given by

The correct option is 15 J/°C, 600 J/kg°C.

Step-by-step Explanation:

We are given the following values from the problem statement:
- Heat energy supplied, $Q=300\text{J}$
- Mass of the sample, $m=25\text{g}=25\times {10}^{-3}\text{kg}=0.025\text{kg}$
- Initial temperature, $T_1={25}^{\circ }\text{C}$
- Final temperature, $T_2={45}^{\circ }\text{C}$

First, let's calculate the rise in temperature ($\mathrm{\Delta }T$):
$\mathrm{\Delta }T=T_2-T_1={45}^{\circ }\text{C}-{25}^{\circ }\text{C}={20}^{\circ }\text{C}$

1. Calculation of Thermal Capacity:
Thermal capacity ($C$) is defined as the amount of heat energy required to raise the temperature of the entire body by 1°C. It is given by the formula:
$C=\frac{Q}{\mathrm{\Delta }T}$
Substituting the given values:
$C=\frac{300\text{J}}{{20}^{\circ }\text{C}}=15\text{J/}{}^{\circ }\text{C}$

2. Calculation of Specific Heat:
Specific heat ($s$) is the heat capacity per unit mass of the material. It is given by the formula:
$s=\frac{C}{m}$
Substituting the values using mass in kg to match the target units:
$s=\frac{15\text{J/}{}^{\circ }\text{C}}{0.025\text{kg}}=600\text{J/kg}{}^{\circ }\text{C}$

Thus, the thermal capacity of the sample is 15 J/°C and the specific heat of the material is 600 J/kg°C.