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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]

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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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Chapter 10: Calorimetry - EXERCISE-1 [Page 233]

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 10
Chapter 10 Calorimetry
EXERCISE-1 | Q 10. (b) | Page 233
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