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

`T_f=57.3\°C`

Step-by-step 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=(m_(cold)T_(cold)+m_(hot)T_(hot))/(m_(cold)+m_(hot))`

Then, we plug in to obtain:

`T_f=(200g*34.5\°C+150g*87.6\C)/(200g+150g)\\\\T_f=57.3\°C`

Best regards!