## Black Principle on Heat Transfer

- Category: Junior High
- Written by fisikastudycenter

* 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:**

m_{1} = 500 gram = 0.5 kg

c_{1} = 385 J/kg°C

ΔT_{1} = 100 − t

**Water:**

m_{2} = 200 gram = 0.2 kg

c_{2} = 4200 J/kg°C

ΔT_{2} = t − 20

**Black principle**

m_{2} x c_{2} x ΔT_{2} = m_{1} x c_{1} x ΔT_{1}

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