Question

29 g of copper pellets are removed from a 300∘C oven and immediately dropped into 90 mL of water at 23 ∘C in an insulated cup. What will the new water temperature be?

Answer 1

Heat Loss by Copper = Heat Taken by water

+Q(water) = -Q(copper)

Q = mcΔT. ΔT = Tf - Ti

Copper, m = 29 g,

Initial temperature of copper, Ti = 300 ºC

C = specific heat of copper = 0.386 J/g ºC

Qcu = 29 g x 0.386 J/g ºC x (Tf – 300) ºC

water, V = 90 mL = 90 g

Initial temperature of water, Ti = 23 ºC

C = specific heat of copper = 4.18 J/g ºC

Qw = 90 g x 4.81 J/g ºC x (Tf – 23) ºC

+Q(water) = -Q(copper)

29 g x 0.386 J/g ºC x (Tf – 300) ºC = 90 g x 4.81 J/g ºC x (Tf – 23) ºC

11.19 (Tf – 300) = -432.9 (Tf – 23)

11.19 Tf + 432.9 Tf = 9956.7 + 3357

444.09 Tf = 13313.7

Tf = 30 ºC