**Example Problems on Temperature, Heat and Heat Transfer and Discussion – ** **Temperature** is a measure of the heat energy stored in an object. High-temperature objects have high heat energy, and vice versa. **Heat** It is defined as the transfer of heat energy from an object with a higher temperature to an object with a lower temperature.

A thermometer is an instrument used to measure the temperature of an object. There are 4 common scales used for thermometers

- Celsius (
^{o}C) - Reamur (
^{o}R) - Fahrenheit (
^{o}F) - Kelvin (K)

There are three types of heat transfer that can occur, including:

- Heat transfer by conduction
- Heat transfer by convection
- Radiation heat transfer

## 1 – 15 Questions on Temperature, Heat and Heat Transfer and Answers

1. Heat of 12 kJ is applied to a piece of metal with a mass of 2500 grams which has a temperature of 30 °C. If the specific heat of the metal is 0.2 calories/gr °C, determine the final temperature of the metal!

**Discussion:**

Diket:

Q = 12 kilojoules = 12000 joules

m = 2500 grams = 2.5 kg

T_{1}= 30 °C

c = 0.2 cal/gr °C = 0.2 x 4200 joules/kg °C = 840 joules/kg °C

T_{2}=…?

**Answer:**

2. 500 grams of ice at 12 °C is heated to a temperature of 2 °C. If the specific heat of ice is 0.5 cal/g °C, determine the amount of heat required, expressed in joules!

**Discussion:**

In the note:

m = 500 grams

T_{1} = 12 °CT_{2} = 2 °C

T = T_{2} T_{1} = 2 (−12 ) = 10 °C

c = 0.5 calories/gr °C

Q = ….?

**Answer:**

Q = m. c . . T

Q = (500)(0,5)(10) = 2500 calories

1 calorie = 4.2 joules

Q = 2500 x 4.2 = 10500 joules

3. 500 grams of ice at 0 °C will be melted until all the ice becomes water at 0 °C. If the specific heat of ice is 0.5 cal/g °C, and the heat of fusion for ice is 80 cal/gr, determine the amount of heat needed, expressed in kilocalories!

**Discussion :**

In the note:

m = 500 grams

L = 80 calories/gr

Q = ….?

**Answer:**

Q = mL

Q = (500)(80) = 40000 calories = 40 kcal

4. 500 grams of ice at 0 °C will be melted to become water at 5 °C. If the specific heat of ice is 0.5 cal/g °C, the heat of fusion for ice is 80 cal/g, and the specific heat of water is 1 cal/g °C, determine how much heat is needed!

**Discussion:**

In the note:

m = 500 grams

c_{liquid} = 1 calorie/gr °C

L_{ice} = 80 calories/gr

Final temperature → 5 °C

Q = …..?

**Answer:**

To turn ice 0°C to 5°C water, there are two processes that must be followed:

→ The process of melting ice 0 °C into water at 0 °C, the heat required is called Q_{1}

Q_{1} = mL_{ice} = (500)(80) = 40000 calories

→ The process of raising the temperature of water 0 °C to 5 °C water, the heat required is called Q_{2}

Q_{2} = mc_{water} T_{water}< = (500) (1)(5) = 2500 calories

Total heat required:

Q = Q_{1} + Q_{2} = 40000 + 2500 = 42500 calories

5. 500 grams of ice at 10 °C will be melted down to become water at 5 °C. If the specific heat of ice is 0.5 cal/g °C, the heat of fusion for ice o is 80 cal/g, and the specific heat of water is 1 cal/g °C, determine how much heat is needed!

**Discussion:**

In the note:

m = 500 grams

c_{ice} = 0.5 calories/gr °C

c_{water} = 1 cal/gr °C

L_{ice} = 80 cal/gr

Final temperature → 5 °C

Q = …..?

**Answer:**

To make ice 10 °C to 5 °C water there are three processes that must be passed:

→ The process for raising the temperature of ice from 10 °C to ice at 0 °C, the heat required is called Q_{1}

Q_{1} = mc_{ice} T_{ice} = (500)(0,5)(10) = 2500 calories

→ The process of melting ice 0 °C into water at 0 °C, the heat required is called Q_{2}

Q_{2} = mL_{ice} = (500)(80) = 40000 calories

→ The process of raising the temperature of water 0 °C to 5 °C water, the heat required is called Q_{3}

Q_{3} = mc_{water} T_{water} = (500)(1)(5) = 2500 calories

Total heat required:

Q = Q_{1} +Q_{2} + Q_{3} = 2500 + 40000 + 2500 = 45000 calories

#### Also Read: Momentum and Impulse Questions and Answers

6. One kilogram of ice has a temperature of – 2 °C, if the melting point of ice = 0 °C, the specific heat of ice = 0.5 cal / g °C, the specific heat of water = 1 cal / g °C, the heat of fusion of ice = 80 cal / g °C and 1 calorie = 4.2 joules, then the heat needed to melt them all is ….

**Discussion:**

Q = Q_{1} + Q_{2}

= m.Δt.c + mk

= 1000 (0 – (-2)) . 0.5 + 1000 . 80

= 1000 + 80,000

= 81,000 calories

= 3.402 x 10^{5} Joule

7. Water with a mass of 100 grams is at a temperature of 20°C heated to a temperature of 80°C. If the specific heat of water is 1 cal/gr°C determine the amount of heat required, express it in units of calories!

**Discussion:**

Question data:

m = 100 grams

c = 1 cal/gr°C

T_{1}= 20°C

T_{2} = 80°C

Heat required:

Q = mxcx T

Q = 100 x 1 x (80−20)

Q = 100 x 60

Q = 6000 calories

8. Water with a mass of 100 grams is heated to a temperature of 20°C until it boils. If the specific heat of water is 4200 J/kg °C determine the amount of heat required, expressed in joules!

**Discussion:**

m = 100 grams = 0.1 kg

c = 4200 J/kg °C

T_{1} = 20°C

T_{2} = 100°C

Heat required:

Q = mxcx T

Q = 0.1 x 4200 x (100−20)

Q = 420 x 80

Q = 33600 joules

9. An ice mass of 200 grams at a temperature of 5°C is heated to a temperature of 1°C, if the specific heat of ice is 0.5 cal/gr°C. Determine how many calories of heat are needed in the process!

**Discussion:**

m = 200 grams

c = 0.5 cal/gr°C

T_{1} = 5°C

T_{2} = 1°C

Heat required:

Q = mxcx T

Q = 200 x 0.5 x [−1−(−5)]

Q = 100 x 4

Q = 400 calories

10. Ice with a mass of 150 grams is heated at 0°C until it completely melts into water at 0°C. Determine the amount of heat required for this process!

**Discussion: **

Question data:

m = 150 grams

L = 80 cal/gr

Heat to melt all the ice:

Q = mx L

Q = 150 x 80

Q = 12000 calories

#### See Also: Example of Fluid Mechanics Problems and Answers

11. Ice with a mass of 250 grams at 5° C is heated until it melts into water at 0°C. If the specific heat of ice is 0.5 cal/gr°C, and the heat of fusion for ice is 80 cal/gr, determine the heat required for the process!

**Discussion:**

Question data:

m = 250 grams

c_{ice} = 0.5 cal/gr°C

L_{ice} = 80 cal/gram

Process 1, raising the temperature of the ice, the heat required:

Q_{1} = mxcx T

Q_{1} = 250 x 0.5 x 5

Q_{1} = 625 calories calories

Process 2, melting all the ice, heat required:

Q_{2} = mx L = 250 x 80 = 20000 calories

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

Q = 625 + 20000

Q = 20625 calories

12. Ice with a mass of 200 grams at 5° C is heated until it melts into water at 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 for ice is 80 cal/gr°C, determine the amount of heat required for the process!

**Discussion:**

Question data:

m_{ice} = m_{water} = 200 grams

c_{ice} = 0.5 cal/gr°C

c_{water} = 1 cal/gr°C

L_{ice} = 80 cal/gram

Process 1, raising the temperature of the ice, the heat required:

Q_{1} = mxcx T

Q_{1} = 200 x 0.5 x 5

Q_{1} = 500 calories

Process 2, melting all the ice, heat required:

Q_{2} = mx L = 200 x 80 = 16000 calories

Process 3, raise water temperature from 0 to 100

Q_{3} = mxcx T

Q_{3} = 200 x 1 x 100

Q_{3} = 30000

The total amount of heat required is the sum of

Q_{1}, Q_{2} and Q_{3}:

Q = 500 + 16000 + 30000

Q = 46500 calories

13. Water at 20°C with a mass of 200 grams is mixed with water at 90°C with a mass of 300 grams. Determine the final temperature of the mixture!

**Discussion:**

Question data:

m_{1} = 200 grams

m_{2} = 300 grams

c_{1} = c_{2} = 1 cal/gr°C

T_{1} = t 20

T_{2} = 90 t

**Answer:**

The principle of exchange of heat/black principle

Q_{free} = Q_{thank}

m_{2} xc_{2} x T_{2} = m_{1} xc_{1} x T_{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

14. A piece of metal with a specific heat of 0.2 cal/gr°C with a mass of 100 grams at a temperature of 30°C is placed in a vessel filled with water at a temperature of 90°C with a mass of 200 grams. If the specific heat of water is 1 cal/gr°C and the effect of the vessel is neglected, determine the final temperature of the metal!

**Discussion:**

Question data:

m_{1} = 100 grams

m_{2} = 200 grams

c_{1} = 0.2 cal/gr°

c_{2} = 1 cal/gr°C

T_{1} = t 30

T_{2} = 90 t

**Answer:**

The principle of exchange of heat/black principle

Qlets = Qaccept

m_{2} xc_{2} x T_{2} = m_{1} xc_{1} x T_{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

15. The following events are related to the heat transfer process:

- 1) iron that is burned at one end, a few moments later the other end is hot.
- 2) the occurrence of land breezes and sea breezes
- 3) sunlight reaches the earth
- 4) the bonfire at a distance of 3 meters is hot
- 5) the exhaust fumes move through the kitchen chimney
- 6) boiled water, the bottom flows up.
- 7) the glass cup is filled with hot water, the outside of the glass feels hot too.
- 8) damp clothes are ironed to dry

Sort the events above based on their relation to heat transfer by conduction, convection and radiation!

**Discussion:**

- 1) iron that is burned at one end, a few moments later the other end feels hot → conduction
- 2) the occurrence of land breezes and sea breezes → convection
- 3) sunlight reaches the earth → radiation
- 4) bonfire at a distance of 3 meters feels hot → radiation
- 5) combustion smoke moves through the kitchen chimney → convection
- 6) boiled water, the bottom flows up → convection
- 7) the glass cup is filled with hot water, the outside of the glass feels hot → conduction
- 8) damp clothes are ironed to dry→ conduction