There was 100 g of water in a glass at a temperature of 20 degrees. 50 g of water was poured into

There was 100 g of water in a glass at a temperature of 20 degrees. 50 g of water was poured into it at a temperature of 80 degrees. What was the temperature after mixing the water.

1. Let’s write down the condition of the problem:

c = 4200 J / kg × ° C

m1 = 100 g = 0.1 kg

t1 = 20 ° C

m2 = 50 g = 0.05 kg

t2 = 80 ° C

t3-?

2. The heat that the second water lost during cooling is equal to that that the first received during heating:

Q1 = Q2

c × m1 × (t3 – t1) = c × m2 × (t2 – t3)

3. Simplify and find t3:

c × m1 × t3 – c × m1 × t1 = c × m2 × t2 – c × m2 × t3

c × m1 × t3 + c × m2 × t3 = c × m2 × t2 + c × m1 × t1

t3 (c (m1 + m2)) = c (m2 × t2 + m1 × t1)

t3 = m2 × t2 + m1 × t1 / m1 + m2

t3 = 0.05 × 80 + 0.1 × 20 / 0.1 + 0.05

t3 = 40 ° C

3. Let’s write down the answer:

Answer: 40 ° C