Question

A quantity of hot water at 91°C and another cold one at 12°C.

Answer 1

Answer:

$m_{cold}=567kg\\\\m_{hot}=233kg$

Explanation:

Hello!

In this case, since equilibrium temperature problems involve the mass, specific heat and temperature change for the substances at different temperatures, we can write:

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

Thus, since we are talking about water and they both have the same specific heat, we can write:

$m_{cold}(T_{eq}-T_{cold})=-m_{hot}(T_{eq}-T_{hot})$

Now, we plug in the temperatures to obtain:

$m_{cold}(35-12)+m_{hot}(35-91)=0\\\\23m_{cold}-56m_{hot}=0$

Next, since the total volume of water is 800 L, since it has a density of 1kg/L, we infer the total mass is 800 kg; that is why we can write a 2x2 system of simultaneous equations:

$\left \{ {{23m_{cold}-56m_{hot}=0} \atop {m_{cold}+m_{hot}=800}} \right.$

Thus, the masses of both cold and hot water turn out:

$m_{cold}=567kg\\\\m_{hot}=233kg$

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