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45 g of water at 50°C in a beaker is cooled when 50 g of copper at 18° C is added to it. The contents are stirred till a final constant temperature is reached. Calculate this final temperature.

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Question

45 g of water at 50°C in a beaker is cooled when 50 g of copper at 18° C is added to it. The contents are stirred till a final constant temperature is reached. Calculate this final temperature. The specific heat capacity of copper is 0.39 J g-1K-1 and that of water is 4.2 J g-1K-1. State the assumption used.

Numerical
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Solution

Mass of water (m1) = 45 g
Temperature of water (T1) = 50°C
Mass of copper (m2) = 50 g
temperature of copper (T2) = 18°C
Final temperature (T) = ?

The specific heat capacity of the copper c2 = 0.39 J/g/K
The specific heat capacity of water c1 = 4.2 J/g/K

m1c1 (T1 - T) = m2 c2 (T - T2

`"T" = ("m"_1"c"_1 "T"_1 + "m"_2 "c"_2"T"_2) /("m"_2 "c"_2 + "m"_1"c"_1 )`

`T = (45 xx 4.2 xx 50 + 50 xx 0. 39 xx 18)/ (45 xx 4.2 + 50 xx 0.39)`

= `9801 / 208.5`

 T = 47°C

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Chapter 11: Calorimetry - EXERCISE-11(A) [Page 271]

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Selina Concise Physics [English] Class 10 ICSE
Chapter 11 Calorimetry
EXERCISE-11(A) | Q 10. | Page 271

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