Question

A $10kW$ drilling machine is used to drill a bore in a small aluminium block of mass $8kg$. How much is the rise in temperature of the block in $2.5\min $, assuming $50\% $ of the power is used up in heating the machine itself and lost to the surrounding. Specific heat of aluminium$ = 0.91L/gK$

Answer 1

Hint: The friction between the drill rod and aluminium block generates heats and the temperature of the block rises. We can find the change in temperature by using the formula $H = mS\Delta T$
Where,
$m$ is mass of the block,
$S$ is specific heat of the block,
$\Delta T$ is a change in temperature.

Complete step by step answer:
The heat generated by the friction will increase the temperature of the block.
It is given in the question that half the power is wasted in heating the drill itself. So, only half of the energy will be used to increase the temperature of the block.
As we know, power is the rate of flow of energy,
$P = \dfrac{E}{t}$
$10000 = \dfrac{E}{{2.5 \times 60}}$
$E = 15 \times {10^5}J$
Amount of energy which will result in rise of temperature $ = \dfrac{{15 \times {{10}^5}}}{2} = 7.5 \times {10^5}J$
Using the formula for heat,
$H = mS\Delta T$
$7.5 \times {10^5} = (8000g)(0.91J/gk)\Delta T$
$\Delta T = 103K$

Additional information:
The specific heat capacity of a substance is the heat capacity of a sample of the substance divided by the mass of the sample. Informally, it is the amount of energy that must be added, in the form of heat, to one unit of mass of the substance in order to cause an increase of one unit in its temperature.
The specific heat often varies with temperature, and is different for each state of matter. Liquid water has one of the highest specific heats among common substances.

Note: Specific heat varies with the change in temperature. In this question, the change in temperature is very small so we can take the specific heat capacity to be constant.