# How much boiling water would you need to raise the bath to body temperature (about 37 ∘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 ∘C). You figure that you can boil water on the stove and pour it into the bath to raise the temperature.

Jan 5, 2018

#### Answer:

$\text{4.8 kg}$

#### Explanation:

The idea here is that the heat given off by the boiling water will be equal to the heat absorbed by the room-temperature sample.

color(blue)(ul(color(black)(q_"absorbed" = -q_"given off")))" " " "color(darkorange)("(*)")

The minus sign is used here because, by convention, heat given off carries a minus sign.

Another assumption that you have to make is that the specific heat of liquid water is constant regardless of the temperature of the liquid water.

In other words, you need to have

${c}_{\text{liquid water at 25"^@"C") = c_ ("liquid water at 100"^@"C}}$

Now, your tool of choice here will be the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot {c}_{\text{liquid water}} \cdot \Delta T}}}$

Here

• $q$ is the heat absorbed or given off
• $m$ is the mass of the sample
• ${c}_{\text{liquid water}}$ is the specific heat of liquid water
• $\Delta T$ is the change in temperature, calculated as the difference between the final temperature and the initial temperature of the sample

So, you know that you have

${q}_{\text{absorbed" = m_1 * c_"liquid water" * DeltaT_"warming}}$

for the room-temperature water, which has

$\Delta {T}_{\text{warming" = 37^@"C" - 25^@"C" = 12^@"C}}$

Similarly, you have

${q}_{\text{given off" = m_2 * c_"liquid water" * DeltaT_"cooling}}$

for the boiling water, which has

$\Delta {T}_{\text{cooling" = 37^@"C" - 100^@"C" = -63^@"C}}$

Use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to get

m_1 * color(red)(cancel(color(black)(c_"liquid water"))) * DeltaT_"warming" = - m_2 * color(red)(cancel(color(black)(c_"liquid water"))) * DeltaT_"cooling"

${m}_{1} \cdot \Delta {T}_{\text{warming" = - m_2 * DeltaT_"cooling}}$

This is equivalent to

${m}_{2} = \left(\Delta {T}_{\text{warming")/(-DeltaT_"cooling}}\right) \cdot {m}_{1}$

Plug in your values to find

m_2 = (12 color(red)(cancel(color(black)(""^@"C"))))/(-(-63color(red)(cancel(color(black)(""^@"C"))))) * "25 kg" = color(darkgreen)(ul(color(black)("4.8 kg")))

Notice that you need the minus sign to cancel out the minus sign coming from the change in temperature.

The answer is rounded to two sig figs.

So, if you add $\text{4.8 kg}$ of liquid water at ${100}^{\circ} \text{C}$ to $\text{25 kg}$ of water at ${25}^{\circ} \text{C}$, you will end up with a mixture that has a final temperature of ${37}^{\circ} \text{C}$.