4. The heat capacity of copper is 0.382 J/gº.C.A 22.6 g sample of copper at 93°...

Question

Question

Answer 1

Answer #1

m(water) = 50.0 g

T(water) = 23.9 oC

C(water) = 4.184 J/goC

m(copper) = 22.6 g

T(copper) = 93.0 oC

C(copper) = 0.382 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by copper = heat gained by water

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

22.6*0.382*(93.0-T) = 50.0*4.184*(T-23.9)

8.6332*(93.0-T) = 209.2*(T-23.9)

802.8876 - 8.6332*T = 209.2*T - 4999.88

T= 26.6386 oC

Answer: 26.6 oC

Answer 2

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