## 9 Common Problems of Heat for Junior

- Category: Junior High
- Written by fisikastudycenter

physics.fisikastudycenter.com - Learning heat for junior hingh school students. Given 9 common problems includes the solutions, the specific heat, the heat of fusion and black principle in heat exchange will be discussed.

** Problem 1**

Water with a mass of 100 gram at a temperature of 20° C then heated to a temperature of 80° C. If the apecific heat of water is 1 cal/gr ° C what is the heat needed in this process show your answer in calorie!

**Answer**

Datal:

m = 100 gram

c = 1 cal/gr°C

T_{1} = 20°C

T_{2} = 80°C

The heat needed:

Q = m x c x Δ T

Q = 100 x 1 x (80−20)

Q = 100 x 60

Q = 6000 calorie

**Problem 2**

100 gram of water at a temperature of 20° C is heated to the temperature of 100°. If the specific heat of water is 4200 J/kg ° C find the heat taken in this process, show in joule!

**Answer**

m = 100 gram = 0.1 kg

c = 4200 J/kg °C

T_{1} = 20°C

T_{2} = 100°C

The heat needed:

Q = m x c x Δ T

Q = 0.1 x 4200 x (100−20)

Q = 420 x 80

Q = 33600 joule

**Problem 3**

200 gram of ice at − 5°C is heated to a temperature of − 1° C, if the specific heat of ice is 0.5 kal/gr ° C determine the heat taken in this process!

**Answer**

m = 200 gram

c = 0.5 kal/gr°C

T_{1} = −5°C

T_{2} = −1°C

The heat:

Q = m x c x Δ T

Q = 200 x 0.5 x [−1−(−5)]

Q = 100 x 4

Q = 400 kalori

**Problem 4**

What is heat needed to melt 150 gram of ice to liquid water of 0°C, use the heat of fusion of ice is 80 cal/g!

**Answer**

Data :

m = 150 gram

L = 80 kal/gr

Heat to melt the ice:

Q = m x L

Q = 150 x 80

Q = 12000 calorie

**Problem 5**

250 gram of ice at a temperature of − 5° C is heated and melts to liquid water of 0°C. If the specific heat of ice is 0.5 cal/gr°C, and the heat of fusion of ice is 80 cal/gr, find the heat taken!

**Answer**

Data:

m = 250 gram

c_{es} = 0.5 kal/gr°C

L_{es} = 80 cal/gram

Step 1, raising the temperature of the ice:

Q_{1} = m x c x ΔT

Q_{1} = 250 x 0,5 x 5

Q_{1} = 625 calorie

Step 2, melting the ice:

Q_{2} = m x L = 250 x 80 = 20000 kalori

The total heat Q_{1} + Q_{2}

Q = 625 + 20000

Q = 20625 calorie

**Problem 6**

200 gram of ice at − 5° C is heated to a liquid water of 100°C. If the specific heat of ice is 0.5 cal/gr°C, the specific heat of water is 1 cal/gr°C and the heat of fusion of ice is 80 cal/gr, find the heat needed in the process!

**Answer**

Datal:

m_{es} = m_{air} = 200 gram

c_{es} = 0.5 cal/gr°C

c_{air} = 1 cal/gr°C

L_{es} = 80 cal/gram

Step 1, raisng ice temperature :

Q_{1} = m x c x ΔT

Q_{1} = 200 x 0.5 x 5

Q_{1} = 500 calorie

Step 2, melting ice:

Q_{2} = m x L = 200 x 80 = 16000 calorie

Step 3, raising the water temperature from 0^{o}C to 100^{o} C

Q_{3} = m x c x ΔT

Q_{3} = 200 x 1 x 100

Q_{3} = 30000 calorie

The total heat Q_{1}, Q_{2} dan Q_{3}:

Q = 500 + 16000 + 30000

Q = 46500 calorie

**Problem 7**

200 gram of water at 20°C is mixed with 300 gram of water at 90°C, calculate the mixture temperature!

**Answer**

Data :

m_{1} = 200 gram

m_{2} = 300 gram

c_{1} = c_{2} = 1 cal/gr°C

ΔT_{1} = t − 20

ΔT_{2} = 90 − t

Black principle of heat exchange:

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

300 x 1 x (90 − t) = 200 x 1 x (t − 20)

27000 − 300t = 200t − 4000

27000 + 4000 = 300t + 200t

31000 = 500t

t = 31000 / 500

t = 62°C

**Problem 8**

A metal at 30°C of specific heat of 0,2 kal/gr°C with 100 gram of mass is dropped into water of 200 gram of mass at a temperature of 90°C. If the specific heat of water is 1 ckal/gr°C calculate the final temperature of the metal, ignore the heat lost in surrounding!

**Answer**

Datal:

m_{1} = 100 gram

m_{2} = 200 gram

c_{1} = 0.2 cal/gr°

c_{2} = 1 cal/gr°C

ΔT_{1} = t − 30

ΔT_{2} = 90 − t

Black principle

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

200 x 1 x (90 − t) = 100 x 0,2 x (t − 30)

18000 − 200t = 20 t − 600

18000 + 600 = 200t + 20t

18600 = 220t

t = 18600 / 220

t = 84,5 °C

**Problem 9**

Given a graph below the temperature vs time:

500 gram of water raises its temperature from C to D. If the specific heat of water is 4200 J/kg°C, find the heat needed in the process, show your answer in kilojoule!

**Answer**

Q = m x c x Δ T

Q = 0.5 x 4200 x 50

Q = 105000 joule

Q = 105 kJ

prep. by:

physics.fisikastudycenter.com