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Question
Calculate the final temperature when 2 kg of cold water at 5°C is mixed with 0.4 kg of hot water at 95°C.
[Specific heat capacity of water = 4.2 J g−1°C−1]
Numerical
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Solution
Given:
Cold Water:
Mass (m2) = 2 kg
Initial Temperature (T2) = 5°C
Hot Water:
Mass (m1) = 0.4 kg
Initial Temperature (T1) = 95°C
Specific heat capacity of water (c): Remains the same for both, so it will cancel out from both sides of our equation.
Final Temperature of the mixture: Let it be T.
m1 × c × (T1 − T) = m2 × c × (T − T2)
Since (c) is on both sides, we can remove it:
0.4 × (95 − T) = 2 × (T − 5)
38 − 0.4 T = 2 T − 10
38 + 10 = 2 T + 0.4 T
48 = 2.4 T
T = \[\frac{48}{2.4}\]
T = 20°C
The final temperature of the mixture will be 20°C.
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