15 Examples of Temperature, Heat and Heat Transfer Problems and their Discussion

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 (oC)
  • Reamur (oR)
  • Fahrenheit (oF)
  • 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

T1= 30 °C

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

T2=…?

Answer:

question of temperature and heat number 1


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

T1 = 12 °CT2 = 2 °C

T = T2 T1 = 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

cliquid = 1 calorie/gr °C

Lice = 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 Q1

Q1 = mLice = (500)(80) = 40000 calories

→ The process of raising the temperature of water 0 °C to 5 °C water, the heat required is called Q2

Q2 = mcwater Twater< = (500) (1)(5) = 2500 calories

Total heat required:

Q = Q1 + Q2 = 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

cice = 0.5 calories/gr °C

cwater = 1 cal/gr °C

Lice = 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 Q1

Q1 = mcice Tice = (500)(0,5)(10) = 2500 calories

→ The process of melting ice 0 °C into water at 0 °C, the heat required is called Q2

Q2 = mLice = (500)(80) = 40000 calories

→ The process of raising the temperature of water 0 °C to 5 °C water, the heat required is called Q3

Q3 = mcwater Twater = (500)(1)(5) = 2500 calories

Total heat required:

Q = Q1 +Q2 + Q3 = 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 ….

question of temperature and heat no. 6


Discussion:

Q = Q1 + Q2

= m.Δt.c + mk

= 1000 (0 – (-2)) . 0.5 + 1000 . 80

= 1000 + 80,000

= 81,000 calories

= 3.402 x 105 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

T1= 20°C

T2 = 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

T1 = 20°C

T2 = 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

T1 = 5°C

T2 = 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

cice = 0.5 cal/gr°C

Lice = 80 cal/gram

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

Q1 = mxcx T

Q1 = 250 x 0.5 x 5

Q1 = 625 calories calories

Process 2, melting all the ice, heat required:
Q2 = mx L = 250 x 80 = 20000 calories

The total heat is Q1 + Q2

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:

mice = mwater = 200 grams

cice = 0.5 cal/gr°C

cwater = 1 cal/gr°C

Lice = 80 cal/gram

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

Q1 = mxcx T

Q1 = 200 x 0.5 x 5

Q1 = 500 calories

Process 2, melting all the ice, heat required:

Q2 = mx L = 200 x 80 = 16000 calories

Process 3, raise water temperature from 0 to 100

Q3 = mxcx T

Q3 = 200 x 1 x 100

Q3 = 30000

The total amount of heat required is the sum of

Q1, Q2 and Q3:

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:

m1 = 200 grams

m2 = 300 grams

c1 = c2 = 1 cal/gr°C

T1 = t 20

T2 = 90 t

Answer:

The principle of exchange of heat/black principle

Qfree = Qthank

m2 xc2 x T2 = m1 xc1 x T1

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:

m1 = 100 grams

m2 = 200 grams

c1 = 0.2 cal/gr°

c2 = 1 cal/gr°C

T1 = t 30

T2 = 90 t

Answer:

The principle of exchange of heat/black principle

Qlets = Qaccept

m2 xc2 x T2 = m1 xc1 x T1

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

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