Question

How much heat energy is gained when 5 kg of water at $20^\circ C\;$ is brought to its boiling point?
A) 1680 kJ
B) 1700 kJ
C) 1720 kJ
D) 1740 kJ

Answer 1

Hint : In this solution, we will use the relation of heat needed to raise the temperature of water by a certain amount. The boiling point of water is $100^\circ C\;$ .

Formula used: In this solution we will be using the following formula,
$\Rightarrow Q = mc\Delta T$ where $Q$ is the energy needed to heat up a liquid by $\Delta T$ temperature difference and $c$ is the specific heat capacity of the liquid

Complete step by step answer
Heat is a form of energy that is transferred from systems having high temperature to low temperature.
We’ve been asked to find the amount of heat energy when 5 kg of water is heated from $20^\circ C\;$ to $100^\circ C\;$ which is the boiling point of water. For this, we can use the formula
$\Rightarrow Q = mc\Delta T$
The specific heat capacity of water is $4.2\,kJ/^\circ C$ . Substituting the value of specific heat capacity and the mass of water $m = 5\,kg$ and $\Delta T = 100 - 20 = 80^\circ C$ , we get
$\Rightarrow Q = (5)4.2(80)$
$\Rightarrow Q = 1680\,kJ$
Hence, we require 1680 kJ of energy to heat 5 kg of water from $20^\circ C\;$ to $100^\circ C\;$ .
So, the correct choice is option (A).

Note
Here we have assumed that all the energy that we provide to the system will be utilized in raising the temperature of the water. However, in practical situations, energy is used in heating up the container and in the air around it so we need to provide some extra heat than what we calculated to bring the water up to boiling temperature. We can only calculate the difference of heat energies of an object but not its absolute value so we can only find the heat energy required to raise the temperature of water by $80^\circ C$ but not the heat energy contained by water at $20^\circ C\;$ or $100^\circ C\;$