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 the final temperature. The specific heat
capacity of copper is 0.39 Jg'Kand that of water
is 4.2 Jg'K. State the assumption used.​

Answer 1

Answer:

Heat gained by copper = m × C × θ

                                    = 50 × 0. 39 × (T − 18)

Heat lost by water = 45 × 4. 2 × (50 − T)

 Heat lost = Heat gained

 45 × 4. 2 × (50 − T) = 50 × 0. 39 × (T − 18)

9450 − 189 T = 19. 5 T − 351

208. 5 T = 9801

Or T = 47°C (Approx)  

 So the final temperature is 47°C (Approx)

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Answer 2

Answer:

hope so it will helpful to you