# Fisika Study Center

## Black Principle on Heat Transfer

physics.fisikastudycenter.com_2021  Black principle on a block of copper and water to find the final temperature.

Problem 1

500 gram of copper at 33.8°C is heated to a temperature of 100° C, what is the heat taken in this process? Use the specific heat of copper is 385 J/kg °C!

Solution

Copper:
m = 500 g = 0.5 kg
c = 4,200 J/kg °C
ΔT = 100°C − 33.8°C = 66.2°C
Q =....?

Q = mcΔT
Q = 0.5(385)(66.2)
Q = 12743.5 joule = 12.74 kJ

Problem 2

500 gram of water at 20°C is heated to a temperature of 33,8° C, if the specific heat of water is 4200 J/kg °C determine the heat taken in this process!

Solution

Water:
m = 500 g = 0.5 kg
c = 4200 J/kg °C
ΔT = 33.8° C − 20°C = 13.8°C
Q =....?

Q = mcΔT
Q = 0.5(4,200)(13.8)
Q = 28980 joule = 28.98 kJ

Note: Remember "." in here is "," in Bahasa, and "," is "." in Bahasa.

Problem 3

A block of copper of specific heat of 385 J/kg°C at a temperature of 100°C with 500 gram of mass is dropped into water of 200 gram of mass at a temperature of 20°C. If the specific heat of water is 4200 J/kg°C what is the final temperature of copper?

Solution

Copper:
m1 = 500 gram = 0.5 kg
c1 = 385 J/kg°C
ΔT1 = 100 − t

Water:
m2 = 200 gram = 0.2 kg
c2 = 4200 J/kg°C
ΔT2 = t − 20

Black principle
m2 x c2 x ΔT2 = m1 x c1 x ΔT1
0.5 x 385 x (100 − t) = 0.2 x 4200 x (t − 20)
19250 − 192.5t = 840 t − 16800
19250 + 16800 = 840t + 192.5t
36050 = 1032.5t

t = 36050/1032.5 = 34.92 °C

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