# How much heat was transferred in this process?

## The amount of boiling water required to raise the temperature of 25.0 kg of water in the bath to body temperature is 4.80 kg. In this process, the heat lost by the boiling water is equal to the heat gained by the room-temperature water

Jul 20, 2016

In the case of this question, there is a discrepancy in the numbers, giving ${q}_{\text{bw" ne q_"rtw}}$ if ${T}_{\text{room" = 25^@ "C}}$, ${T}_{b} = {100}^{\circ} \text{C}$, and ${T}_{\text{body" = 37^@ "C}}$.

So I have assumed ${T}_{\text{room" ~~ 24.8^@ "C}}$.

Body temperature is ${37}^{\circ} \text{C}$, boiling temperature for water is ${100}^{\circ} \text{C}$, and room temperature is typically about ${25}^{\circ} \text{C}$. However, I will be using ${T}_{\text{room" ~~ 24.8^@ "C}}$.

"How much heat was transferred from $\text{4.80 kg}$ of ${100}^{\circ} \text{C}$ water when it was poured into $\text{25.0 kg}$ of ${24.8}^{\circ} \text{C}$ water to bring it up to ${37}^{\circ} \text{C}$?"

As usual, you would use the heat flow equation:

$\setminus m a t h b f \left(q = m s \Delta T\right)$,

where:

• $q$ is the heat flow in $\text{J}$.
• $m$ is the mass of one thing in $\text{g}$.
• $s$ is the specific heat capacity in $\text{J/g"cdot""^@ "C}$.
• $\Delta T$ is the change in temperature due to the heat flow. You can use $\text{^@ "C}$, but $\text{K}$ is OK too, since their intervals are identical.

You also know that the heat lost by the boiling water (${q}_{\text{bw}}$) is claimed to be equal to the heat gained by the room-temperature water (${q}_{\text{rtw}}$), so what you have is supposedly:

$\setminus m a t h b f \left(- {q}_{\text{bw" + q_"rtw}} = 0\right)$,

where the signs account for the gain or loss of heat (negative is loss, positive is gain, with respect to the body of water in question).

So, based on our sign conventions, the numbers we get from each calculation must be identical, as long as we write ${T}_{f} > {T}_{i}$, and our specific heat capacities are from consistent sources.

Also, the water stops changing temperature at thermal equilibrium, where both bodies of water become the same temperature (${37}^{\circ} \text{C}$). Therefore, ${T}_{f}^{\text{bw" = T_f^"rtw}}$, necessarily.

The specific heat capacity of boiling water is NOT $4.184$. It is actually about $\text{4.219 J/g"cdot""^@ "C}$, according to this source. You should use the numbers you have in your book, however, to be consistent.

The heat transferred from the boiling water was:

color(blue)(q_"bw") = m_"bw"s_"bw"DeltaT_"bw"

= m_"bw"s_"bw"(T_f^"bw" - T_i^"bw")

$= \text{4800 g" xx "4.219 J/g"cdot""^@ "C" xx |37^@ "C" - 100^@ "C} |$

$\implies$ $\textcolor{b l u e}{\text{1275.83 kJ}}$

And the heat transferred into the room-temperature water was:

color(blue)(q_"rtw") = m_"rtw"s_"rtw"DeltaT_"rtw"

= m_"rtw"s_"rtw"(T_f^"rtw" - T_i^"rtw")

$= \text{25000 g" xx "4.184 J/g"cdot""^@ "C" xx |37^@ "C" - 24.8^@ "C} |$

$\implies$ $\textcolor{b l u e}{\text{1276.12 kJ}}$ $\approx$ $\text{1275.83 kJ}$

(0.02% error.)

If I had used ${T}_{\text{room" = 25^@ "C}}$, then with 3 sig figs, $1280 - 1260 = \text{20 kJ}$ of heat is unaccounted for, which is significant (~~1.57% error).

Either this is a realistic problem (where some heat was transferred out to the atmosphere), or the room-temperature water was actually more like ${24.8}^{\circ} \text{C}$.

The number for room temperature would probably account for it, since it's not that far off from ${25}^{\circ} \text{C}$. Actually, some of my university's labs are at about ${22}^{\circ} \text{C}$.

With room temperature at ${24.803}^{\circ} \text{C}$ rather than ${25}^{\circ} \text{C}$:

color(blue)(|q_"rtw"|) = |q_"bw"|

$=$ $\textcolor{b l u e}{\text{1275.83 kJ}}$

~~ "25000 g" xx "4.184 J/g"cdot""^@ "C" xx |37^@ "C" - color(blue)(24.803^@ "C")|