A 300g metal bar at 23 degrees C is placed in a oven. 23,000 J of the heat added to the bar causes the temperature of the bar to increase up to 193.37 degrees C. What is the specific heat(c) of the bar?
Show your work!
Thanks!
Show your work!
Thanks!
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q = (m)(Sc)(ΔT)
And rearrange to solve for specific heat:
Sc = q/(m)(ΔT)
q = 23000 J
m = 300 g
ΔT = Tf - Ti = 193.37°C - 23°C = 170.37°C
Sc = 23000 J/ (300 g)(170.37°C)
Sc = 23000 J/ 51111 g*°C
Sc = 0.45 J/g*°C
And rearrange to solve for specific heat:
Sc = q/(m)(ΔT)
q = 23000 J
m = 300 g
ΔT = Tf - Ti = 193.37°C - 23°C = 170.37°C
Sc = 23000 J/ (300 g)(170.37°C)
Sc = 23000 J/ 51111 g*°C
Sc = 0.45 J/g*°C