Question

22 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 25 ∘C in an insulated cup. What will the new water temperature be?

Answer 1

Specific heat capacity of water = 4.2 J /cal / g

for copper = 0.385  J /cal / g

now

let the final temperature be T

Heat lost = Heat gained

Heat lost by copper = 22 x 0.385 x (300 - T)

heat gained by water = 120 x 4.2 x (T -25)

22 x 0.385 x (300 - T) = 120 x 4.2 x (T -25)

8.47 (300 - T) = 504 (T -25)

2541 - 8.47 T = 504 T - 12600

T =29.545 C