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.

Answer 1

#"4.8 kg"#

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_ ("liquid water at 25"^@"C") = c_ ("liquid water at 100"^@"C")#

Now, your tool of choice here will be the equation

#color(blue)(ul(color(black)(q = m * c_"liquid water" * DeltaT)))#

Here

So, you know that you have

#q_"absorbed" = m_1 * c_"liquid water" * DeltaT_"warming"#

for the room-temperature water, which has

#DeltaT_"warming" = 37^@"C" - 25^@"C" = 12^@"C"#

Similarly, you have

#q_"given off" = m_2 * c_"liquid water" * DeltaT_"cooling"#

for the boiling water, which has

#DeltaT_"cooling" = 37^@"C" - 100^@"C" = -63^@"C"#
Use equation #color(darkorange)("(*)")# 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 * DeltaT_"warming" = - m_2 * DeltaT_"cooling"#

This is equivalent to

#m_2 = (DeltaT_"warming")/(-DeltaT_"cooling") * 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 #"4.8 kg"# of liquid water at #100^@"C"# to #"25 kg"# of water at #25^@"C"#, you will end up with a mixture that has a final temperature of #37^@"C"#.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

You would need approximately twice the volume of water as the volume of the bath.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7