Explanation:
There are two containers of water, one colder and one warmer.
The heat gained by the colder water must equal the heat lost by the warmer water, so the net heat transfer is 0.
∑q = 0
m₁C₁ΔT₁ + m₂C₂ΔT₂ = 0
Since both containers have water with the same density and the same heat capacity, we can simplify this to:
V₁ΔT₁ + V₂ΔT₂ = 0
In the first example:
240 (52 − T₁) + 260 (52 − T₂) = 0
In the second example:
120 (46 − T₁) + 180 (46 − T₂) = 0
Two equations, two unknowns. Start by simplifying both equations:
26000 − 240 T₁ − 260 T₂ = 0
13800 − 120 T₁ − 180 T₂ = 0
Double the second equation:
27600 − 240 T₁ − 360 T₂ = 0
Subtract the first equation:
1600 − 100 T₂ = 0
T₂ = 16
Plug into any either equation to find T₁:
T₁ = 91
The first container has a temperature of 91°C, and the second container has a temperature of 16°C.