Question

A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the water? Group of answer choices

Answer 1

Answer:
`T_f=33.85\°C`

Step-by-step explanation:

Hello,

In this case, we can write the following relationship, explaining that the lost by the hot water is gained by the cold water:

`Q_(hot,W)=-Q_(cold,W)`

Which in terms of mass, specific heat and temperatures, we have:

`m_(hot,W)Cp_(W)(T_f-T_(hot,W))=-m_(cold,W)Cp_(W)(T_f-T_(cold,W))`

Whereas the specific heat of water is cancelled out to obtain the following temperature, considering that the density of water is 1 kg/L:

`T_f=(m_(hot,W)T_(hot,W)+m_(cold,W)T_(cold,W))/(m_(hot,W)+m_(cold,W))\\\\T_f=(5.0kg*80\°C+60kg*30\°C)/(5.0kg+60kg) \\\\T_f=33.85\°C`

Regards.