Heat Transfer

Aug 2018
Please help clarify which is the correct answer and or method to the following problem:

Ideally, when a thermometer is used to measure the temperature of an object, the temperature of the object itself should not change. However, if a significant amount of heat flows from the object to the thermometer, the temperature will change. A thermometer has a mass of 31 g, a specific heat capacity of 815 J/kg·°C, and a temperature of 12°C. It is immersed in 119 g of water, and the final temperature of the water and thermometer is 41.5°C. What was the temperature of the water before the insertion of the thermometer?

The answer posted by the instructor is 43 degrees C.

I am confused because I'm getting conflicting instructions from different sources, or maybe I'm just not understanding the full concept.

The instructor's way uses:
m1c1(Tf-Ti) = m2c2(Tf-Ti)

with this I got 40 degrees C.

Another source instructed to use:
m1c1(Tf-Ti) = m2c2(Ti-Tf)

with this I got 43 degrees C.

I do understand that Q=m1c1(Tf-Ti).
So, is the posted answer incorrect, or is the instructor's method incorrect, or is it just my calculation? Thanks and much appreciated! The attached image shows my work.


Apr 2017
Pheeew ...Look at all those equations and numbers in your work , makes my eyes glaze over .... No need for it , just walk through the question step by step.....

The thermometer has mass 31 gm ,heat capacity 815/kg = 25.265J/C ...

Initial temp of thermometer is 12 final temp is 41.5 ..temp change is 29.5

That means the thermometer extracts 29.5 x 25.265 J = 745.3175J from the water.

thermal capacity of water 4.185 x119 = 498J/C

so how much will temp of water change if 745J is taken ???

745/ 498 = 1.5C ....gives the answer 43C
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Oct 2017
Oz's answer is great. However, let's get your confusion with the equations sorted out as well.

TL;DR answer:
There is a negative sign being introduced by the fact that the water loses heat whereas the thermometer gains heat. Introducing this sign causes the \(\displaystyle T_f\) and \(\displaystyle T_i\) to swap, giving you the correct answer of 43 deg. C.

Longer answer:

Generally, for sensible heat transfer:
\(\displaystyle \Delta Q = m c \Delta T\)

The only difference between the two equations you provide is which order the temperatures are: \(\displaystyle (T_f - T_i)\) or \(\displaystyle (T_i - T_f)\).

Therefore, the question boils down to (pun intended!) whether you think the thermometer is increasing or decreasing its temperature.

You should think carefully about the heat transfer happening between the objects in your problem and resulting change in temperatures of the heat reservoirs. If something increases its temperature, \(\displaystyle \Delta Q\) is positive. If something decreases its temperature, \(\displaystyle \Delta Q\) is negative.

What you have is conservation of energy as heat flows from one object to the other. The water (1) loses heat to the thermometer (2) via sensible heat transfer, so if we look at the equation for water:

\(\displaystyle Q_1 = m_1 c_1 (T_{1,f} - T_{1,i})\)

we expect the final temperature of the water to be lower than the initial temperature and the result is negative (i.e. \(\displaystyle Q_1 < 0\)). Since the thermometer gains the heat lost from the water, we have

\(\displaystyle Q_2 = - Q_1\)


\(\displaystyle m_2 c_2 (T_{2,f} - T_{2,i}) = - m_1 c_1 (T_{1,f} - T_{1,i})\)

\(\displaystyle m_2 c_2 (T_{2,f} - T_{2,i}) = m_1 c_1 (T_{1,i} - T_{1,f})\)
Last edited:
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Aug 2018
benit13, thank you! That makes a lot of sense now.

oz93666, I will keep in mind that method. Thanks.