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
$\displaystyle Q_2 = - Q_1$
Therefore,
$\displaystyle m_2 c_2 (T_{2,f} - T_{2,i}) = - m_1 c_1 (T_{1,f} - T_{1,i})$
$\displaystyle m_2 c_2 (T_{2,f} - T_{2,i}) = m_1 c_1 (T_{1,i} - T_{1,f})$
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Last edited by benit13; Aug 24th 2018 at 06:29 AM.
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