Question

The radiator of a car contains 20L water. If the motor supplies $2 \times {10^5}$ Calories heat to it, then rise in its temperature will be
A. $1000^\circ C$
B. $100^\circ C$
C. $20^\circ C$
D. $10^\circ C$

Answer 1

Hint Heat is always transferred from a body at higher temperature to a body at lower temperature. This heat is used to raise the temperature of a substance to a threshold level. After that threshold level, the body changes its state. For water, this temperature is at 100 C, after which heat will be used to change its state from liquid to steam. After water has become vapour, further heat will be used to heat up the steam.

Complete step by step solution
We know that
$Q = mc\Delta T \\\$
Where m is the mass, c is the specific heat and $\Delta T$ is the temperature difference.

Given that radiator contains 20 L of water and $Q = 2 \times {10^5}Cal = 200Kcal$
The mass of the water present will be ${\rho _{water}} \times V$
${\rho _{water}} = 1000{\text{ }}kg/{m^3} \\\ V = 20L = 0.02{m^3} \\\ mass,m = 0.02 \times 1000 = 20kg = 20 \times {10^3}grams \\\$
$2 \times {10^5}Cal = 20 \times {10^3} \times 1 \times \Delta T \\\ \Delta T = \dfrac{{2 \times {{10}^5}Cal}}{{20 \times {{10}^3}}} = {10^o}C \\\$

And specific heat of water is $1cal{g^{ - 1}}^o{C^{ - 1}}$

Hence the temperature rise is of $10^\circ C$ that is option (D)

Note: We know that $Q = mc\Delta T$, we substitute the values in the equation and get $\Delta T$ that is $10^\circ C$. The unit conversion is very important as volume is given in L should be converted into ${m^3}$ and specific heat of water is $1cal{g^{ - 1}}^o{C^{ - 1}}$. So the mass obtained in kg is to be converted into grams.