Question

3000 grams of water (at a temperature of 20 °C) is mixed with 1000 grams of hot

water(at a temperature of 80 °C). Assuming no heat energy is exchanged with the
environment, what is the final equilibrium temperature? Specific heat of water is
4.19 J/g °C.
35 °C
40 °C
45 °C
50 °C

Answer 1

35°C

Step-by-step explanation:

Heat gained by matter is:

q = mCΔT

where m is mass, C is specific heat capacity, and ΔT is change in temperature.

There is no total heat transfer, so:

0 = m₁C₁ΔT₁ + m₂C₂ΔT₂

The specific heat capacity is the same for both, so divide by C:

0 = m₁ΔT₁ + m₂ΔT₂

Plug in values:

0 = 3000 (T − 20) + 1000 (T − 80)

Solve for T:

0 = 3000T − 60,000 + 1000T − 80,000

140,000 = 4000T

T = 35

The final temperature is 35°C.