# How much heat was transferred in this process?

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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

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

##### 1 Answer

In the case of this question, there is a discrepancy in the numbers, giving

So I have assumed

**Body temperature** is **boiling temperature** for water is **room temperature** is typically about

So, the question is asking:

"How much heat was transferred from

#"4.80 kg"# of#100^@ "C"# water when it was poured into#"25.0 kg"# of#24.8^@ "C"# water to bring it up to#37^@ "C"# ?"

As usual, you would use the **heat flow equation**:

#\mathbf(q = msDeltaT)# ,where:

#q# is theheat flowin#"J"# .#m# is themassof one thing in#"g"# .#s# is thespecific heat capacityin#"J/g"cdot""^@ "C"# .#DeltaT# is thechange in temperaturedue to the heat flow. You can use#""^@ "C"# , but#"K"# is OK too, since their intervals are identical.

You also know that the heat lost by the boiling water (** equal** to the heat gained by the room-temperature water (

#\mathbf(-q_"bw" + q_"rtw" = 0)# ,where the

signsaccount 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 (

The specific heat capacity of ** boiling** water is NOT

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")#

#= "4800 g" xx "4.219 J/g"cdot""^@ "C" xx |37^@ "C" - 100^@ "C"|#

#=># #color(blue)("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")#

#= "25000 g" xx "4.184 J/g"cdot""^@ "C" xx |37^@ "C" - 24.8^@ "C"|#

#=># #color(blue)("1276.12 kJ")# #~~# #"1275.83 kJ"# (

#0.02%# error.)

If I had used

Either this is a realistic problem (where some heat was transferred out to the atmosphere), or the room-temperature water was actually more like

The number for room temperature would probably account for it, since it's not that far off from

With room temperature at

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

#=# #color(blue)("1275.83 kJ")#

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