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:$T_f=33.85\°C$

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:

$T_f=\frac{m_{hot,W}T_{hot,W}+m_{cold,W}T_{cold,W}}{m_{hot,W}+m_{cold,W}}\\\\T_f=\frac{5.0kg*80\°C+60kg*30\°C}{5.0kg+60kg} \\\\T_f=33.85\°C$

Regards.