Mixed 6 kg of water at 42 ° C, 4 kg of water at 72 ° C and 20 kg of water at an unknown temperature.

Mixed 6 kg of water at 42 ° C, 4 kg of water at 72 ° C and 20 kg of water at an unknown temperature. Determine this temperature if the steady-state temperature is 30 ° C.

Data: mw1 – mass of the first part of the mixture (mw1 = 6 kg); tv1 – initial temperature of the first part of the mixture (tv1 = 42 ºС); mw2 is the mass of the second part of the mixture (mw2 = 4 kg); tv2 – initial temperature of the second part of the mixture (tv2 = 72 ºС); mw3 is the mass of the third part of the mixture (mw3 = 30 kg); tр – equilibrium temperature (tр = 30 ºС).

To find out the initial temperature of the third part of the mixture, consider the equality:

Cw * mw1 * (tv1 – tp) + Cw * mw2 * (tv2 – tr) = Cw * mw3 * (tr – tv3).

mv1 * (tv1 – tr) + mv2 * (tv2 – tr) = mv3 * (tr – tv3).

Payment:

6 * (42 – 30) + 4 * (72 – 30) = 20 * (30 – tv3).

72 + 168 = 600 – 20tv3.

20tw3 = 600 – 72 – 168.

20tw3 = 360.

tv3 = 360/20 = 18 ºС.

Answer: The initial temperature of the third part of the mixture was 18 ° C.