Question

Attempt 2 - Question 5 of 5 > A hot lump of 40.6 g of copper at an initial temperature of 81.2 °C is placed in 50.0 mL H,O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/(g°C)? Assume no heat is lost to surroundings. Tanal = TOOLS x10

Answer 1

Density of water is 1.00 g/mL and volume is 50.0 mL.

So, mass of water,

m(water) = 50.0 g

T(water) = 25.0 oC

C(water) = 4.184 J/goC

m(Cu) = 40.6 g

T(Cu) = 81.2 oC

C(Cu) = 0.385 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by Cu = heat gained by water

m(Cu)*C(Cu)*(T(Cu)-T) = m(water)*C(water)*(T-T(water))

40.6*0.385*(81.2-T) = 50.0*4.184*(T-25.0)

15.631*(81.2-T) = 209.2*(T-25.0)

1269.2372 - 15.631*T = 209.2*T - 5230

T= 28.9072 oC

Answer: 28.9 oC