The glass contained 100 g of water at a temperature of 70 ° C. Added another 50 g of water at a temperature

The glass contained 100 g of water at a temperature of 70 ° C. Added another 50 g of water at a temperature of -20 degrees C. What will be the temperature of the mixture?

Given:
m1 = 100g = 0.1kg,
m2 = 50g = 0.05kg,
t1 = 70 ° C,
t2 = 20 ° C,
Let’s designate: c – specific heat capacity of water,
Find t3 -?
Let’s write the heat balance equation:
Q1 = Q2;
Q1 = m1 * c * (t1 – t3);
Q2 = m2 * c * (t3 – t2);
m1 * c * (t1 – t3) = m2 * c * (t3 – t2);
0.1 * (70 – t3) = 0.05 * (t3 – 20);
140 – 2 * t3 = t3 – 20;
3 * t3 = 160
t3 = 53.3 ° C.