In a calorimeter, 100 g of water was mixed, taken at a temperature of 40C and 100C, and 50 g of water was added

In a calorimeter, 100 g of water was mixed, taken at a temperature of 40C and 100C, and 50 g of water was added at a temperature of 60C and 80C. what is the temperature of the mixture.

m1 = 100 g = 0.1 kg.

t1 = 40 ° C.

t2 = 100 ° C.

m2 = 50 g = 0.05 kg.

t3 = 60 ° C.

t4 = 80 ° C.

t -?

Find the temperature t “at the first mixing.

C * m1 * (t “- t1) = C * m1 * (t2 – t”).

t “- t1 = t2 – t”.

t “= (t2 + t1) / 2.

t “= (100 ° C + 40 ° C) / 2 = 70 ° C.

Now we have obtained water of mass m1 “= 2 * m1 with temperature t” = 70 ° C.

Let us find the temperature t1 “when mixing water with masses m2 = 0.05 kg and temperatures t3 = 60 ° C and t4 = 80 ° C.

С * m3 * (t1 “- t3) = С * m4 * (t4 – t1”).

t1 “- t3 = t4 – t1”.

t1 “= (t3 + t4) / 2.

t1 “= (60 ° C + 80 ° C) / 2 = 70 ° C.

The result was water with a mass of m3 “= 2 * m3 with a temperature of t1” = 70 ° C.

If you mix water with the same temperature t “= 70 ° C and t1” = 70 ° C, you get water of the same temperature.

Answer: when mixing water, a temperature of t = 70 ° C will be established.