Question

The specific heat C of a solid at low temperature shows temperature dependence according to the relation \[C = D{T^3}\], where D is a constant and T is the temperature in Kelvin. A piece of this solid of mass m kg is taken and its temperature is raised from 20K to 30K. The amount of heat required in the process in energy units is:
A. \[5 \times {10^2}\,Dm\]
B. \[\left( {33/4} \right) \times {10^4}\,Dm\]
C. \[\left( {65/4} \right) \times {10^4}\,Dm\]
D. \[\left( {5/4} \right) \times {10^4}\,Dm\]

Answer 1

Hint: Use the formula for heat lost or gained by the substance of mass m and specific heat capacity C. Integrate the equation with given value of specific heat from given temperature range.

Formula used:
The small change in heat due to small change in temperature is expressed as,
\[dQ = mCdT\]
Here, m is the mass of the substance and C is the specific heat capacity of the substance.

Complete step by step answer:
We know that the amount of heat required by the substance to raise the temperature is,
\[Q = \int {dQ} \] …… (1)

We know that the heat lost or gain by the material of specific heat capacity \[C\] is given as,
\[dQ = mCdT\]
 We substitute the above equation in equation (1).
\[Q = \int\limits_{T = 20K}^{T = 30K} {mCdT} \]
We have given, the temperature dependence of specific heat of the given material is \[C = D{T^3}\].

Therefore, we have to substitute \[C = D{T^3}\] in the above equation.
 \[Q = \int\limits_{T = 20K}^{T = 30K} {m\left( {D{T^3}} \right)dT} \]
\[ \Rightarrow Q = mD\int\limits_{T = 20K}^{T = 30K} {{T^3}dT} \]
\[ \Rightarrow Q = mD\left( {\dfrac{{{T^4}}}{4}} \right)_{20K}^{30K}\]
\[ \Rightarrow Q = mD\left( {\dfrac{{{{\left( {30} \right)}^4}}}{4} - \dfrac{{{{\left( {20} \right)}^4}}}{4}} \right)\]
\[ \therefore Q = \dfrac{{65}}{4} \times {10^4}mD\]

So, the correct answer is option (C).

Note: The integration of the above used function is, \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1 \pm }} + C} \], where C is the constant of integration. The specific heat capacity of the material is temperature dependent. For most of the materials, it increases with increase in temperature. To solve the questions for which students have to solve the integrations, students should remember the important integration formulae.