Question

1. Mixing Water at Two Temperatures

Answer 1

Answer:

$T_f=72.5\°C$

Explanation:

Hello!

In this case, since this a problem in which the cold water is heated by the hot water, we can write:

$Q_{hot}+Q_{cold}=0$

Thus, by plugging in the mass, specific heat and temperatures, we obtain:

$m_{hot}C_{hot}(T_f-T_{hot})+m_{cold}C_{cold}(T_f-T_{cold})=0$

Now, we can also write:

$m_{hot}(T_f-T_{hot})+m_{cold}(T_f-T_{cold})=0$

Then, after applying some algebra, it is possible to obtain:

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

If we plug in, we obtain:

$T_f=\frac{350.0g*95.0\°C+150.0g*20.0\°C}{350.0g+150.0g}$

$T_f=72.5\°C$

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