Question

200g of water at 34.5°C are added to 150g of water at 87.6°C. What is the final temperature of the mixture?

Answer 1

Answer:

$T_f=57.3\°C$

Explanation:

Hello there!

In this case, for this calorimetry problem, it is possible to realize that the hot water at 87.6 °C is cooled down whereas the cold water at 34.5 °C is heated up, according to:

$Q_{cold}=-Q_{hot}$

Which in terms of mass, specific heat (cancelled out because they have the same value for being water) and temperature difference, is:

$m_{cold}C_{cold}(T_f-T_{cold})=-m_{hot}C_{hot}(T_f-T_{hot})\\\\m_{cold}(T_f-T_{cold})=-m_{hot}(T_f-T_{hot})$

Thus, solving for the final temperature, we obtain:

$T_f=\frac{m_{cold}T_{cold}+m_{hot}T_{hot}}{m_{cold}+m_{hot}}$

Then, we plug in to obtain:

$T_f=\frac{200g*34.5\°C+150g*87.6\C}{200g+150g}\\\\T_f=57.3\°C$

Best regards!