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 J - K-
and that of water is 4.2 J g-1 K-1. 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)

i hope it will help you

regards

Answer 2

Answer:

47.007°C

Formula used is given in the first step of the attachment. m1 is the mass of water, c1 is the sp. heat capacity of water, t1 is the temperature of water. t is the final temperature here. m2, c2 and t2,and the mass , sp. heat capacity and temperature of copper respectively .