Question

How much boiling water would you need to raise the bath to body temperature (about $37{}^\circ C$)? Assume that no heat is transferred to the surrounding environment. Express your answer to two significant figures and include the appropriate units.
You fill your bathtub with 25 kg of room-temperature water (about $25{}^\circ C$). You figure that you can boil water on the stove and pour it into the bath to raise the temperature.

Answer 1

Hint: This question is based on the concept of heat capacity. The amount of heat that is required by a substance to produce a unit change in its temperature is known as the heat capacity of a substance or thermal capacity. Its SI unit is joules per kelvin or J/K.
We get the specific heat capacity of a substance when we divide the heat capacity of the substance by its mass. Its SI unit is joules per kelvin and kilogram or J/kg.K.

Complete answer:
We know that to change the temperature of one unit of mass of a substance by one degree, the amount of energy that is required is equivalent to the specific heat capacity of the substance.
Now, from the definition, we can see that specific heat capacity is
\[c=\dfrac{C}{m}=\dfrac{Q}{\Delta T\times m}\]
This equation can be re-written as
\[Q=mc\Delta T\]
Where c is the specific heat capacity, C is the heat capacity (J/K) of the substance, Q is the heat energy (joules), m is the mass of the substance (kg), and $\Delta T$ is the change in the temperature (K).
Now, heat energy absorbed by the water at room temperature is given by
\[{{Q}_{abs}}={{m}_{1}}c\Delta {{T}_{H}}\]
Where, \[{{Q}_{abs}}\] is the heat absorbed,
\[{{m}_{1}}\] is the mass of water at room temperature = 25 kg
And \[\Delta {{T}_{H}}\] is the increase in the temperature of water = 37-25 = 12${}^\circ C$.
Similarly, the heat energy released by the boiling water is given by
\[{{Q}_{rel}}={{m}_{2}}c\Delta {{T}_{C}}\]
Where, \[{{Q}_{rel}}\] is the heat released,
\[{{m}_{2}}\] is the mass of boiling water
And \[\Delta {{T}_{C}}\] is the decrease in the temperature of boiling water = 100-37 = 63${}^\circ C$.
Since, in this question, it is given to us that when boiling water is added to the water at room temperature, all the heat imparted by the boiling water will be absorbed by the water at room temperature. So,
\[{{Q}_{abs}}={{Q}_{rel}}\]
By using the formula and substituting values we get
\[\begin{align}
  & {{m}_{1}}c\Delta {{T}_{H}}={{m}_{2}}c\Delta {{T}_{C}} \\
 & {{m}_{1}}\Delta {{T}_{H}}={{m}_{2}}\Delta {{T}_{C}} \\
 & {{m}_{2}}=\dfrac{\Delta {{T}_{H}}}{\Delta {{T}_{C}}}{{m}_{1}} \\
 & {{m}_{2}}=\dfrac{12}{63}\times 25 \\
 & {{m}_{2}}\cong 4.762kg \\
\end{align}\]
So, the addition of approximately 4.8 kg of boiling water (100${}^\circ C$) to 25 kg of water at room temperature (25${}^\circ C$) will raise its temperature to body temperature (37${}^\circ C$).

Note:
It must be noted that we have made two assumptions in this question.
1. There is no loss of heat energy upon transfer of heat from boiling water to room temperature water.
2. The specific heat of water is the same at different temperatures, i.e., the specific heat of boiling water is the same as the specific heat of water at room temperature.