Temperature and heat – problems and solutions

1. On a thermometer X, the freezing point of water at -30o and the boiling point of water at 90o. 60OX = ….. oC.

Known :

The freezing point of water = -30o

The boiling point of water = 90o

Wanted : 60oX = ….. oC

Solution :

On the Fahrenheit scale, the freezing point of water is 32oF and the boiling point of water is 212oF. Between the freezing point and the boiling point, 212o – 32o = 180o.

On the Celsius scale, the freezing point of water is 0oC and the boiling point of water is 100oC. Between the freezing point and the boiling point, 100o – 0o = 100o.

On the X scale, the freezing point of water is -30oX and the boiling point of water is 90oX. Between the freezing point and the boiling point, 90o – (-30o) = 90o + 30o = 120o.

Change the X scale to the Celsius scale :

Temperature and heat – problems and solutions 1

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Thermal expansion

2. A metal rod heated from 30oC to 80oC. The final length of the rod is 115 cm. The coefficient of linear expansion is 3.10-3 oC-1. What is the initial length of the metal rod?

Known :

The initial temperature (T1) = 30oC

The final temperature (T2) = 80oC

The change in temperature (ΔT) = 80oC – 30oC = 50oC

The coefficient of linear expansion (α) = 3.10-3 oC-1

The final length of the metal (L) = 115 cm

Wanted : The initial length of the metal rod (Lo)

Solution :

The equation of the linear expansion :

L = Lo + ΔL

L = Lo + α Lo ΔT

L = Lo (1 + α ΔT)

115 = Lo (1 + 3.10-3.50)

115 = Lo (1 + 150.10-3)

115 = Lo (1 + 0.15)

115 = Lo (1.15)

Lo = 115 / 1.15

Lo = 100 cm

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3. The initial length of a brass rod is 40 cm. After heated, the final length of the brass is 40.04 cm and the final temperature is 80oC. If the coefficient of linear expansion of the brass is 2.0 x 10-5 oC1, what is the initial temperature of the brass rod.

Known :

The final temperature (T2) = 80oC

The initial length (Lo) = 40 cm

The final length (L) = 40.04 cm

The increase in length (ΔL) = 40.04 cm – 40 cm = 0.04 cm

The coefficient of linear expansion (α) = 2.0 x 10-5 oC-1

Solution : Initial temperature (T1)

Solution ;

The equation of the linear expansion :

L = Lo + α Lo ΔT

L – Lo = α Lo ΔT

ΔL = α Lo ΔT

ΔL = α Lo (T2 – T1)

0.04 = (2.0 x 10-5)(40)(80 – T1)

0..04 = (80 x 10-5)(80 – T1)

0.04 = 0.0008 (80 – T1)

0.04 = 0.064 – 0.0008 T1

0..0008 T1 = 0.064 – 0.040

0.0008 T1 = 0.024

T1 = 30oC

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Heat transfer conduction

4. Two metal rods with the same size but different type, as shown in figure below. The thermal conductivity of metal I = 4 times the thermal conductivity of metal II. What is the temperature between both metals.

Known :

The size of both rods is the same.Temperature and heat – problems and solutions 2

The thermal conductivity of metal I = 4k

The thermal conductivity of metal II = k

The temperature of the one end of metal I = 500 C

The temperature of the one end of metal II = 00 C

Wanted: The temperature between both the metal rods

Solution :

The equation of the heat conduction :

Temperature and heat – problems and solutions 3

Q/t = the rate of heat conduction, k = thermal conductivity, A = the cross-sectional area, T1-T2 = the change in temperature, l = the length of the rod.

The temperature between both rods :

Temperature and heat – problems and solutions 4

The temperature at the center between both the metals rods is 40oC.

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5.

(1) Conductivity of metal

(2) The difference of temperature

(3) The length of metal

(4) Mass of metal

Factors that determine the rate of heat conduction on metals are…..

Solution :

Based on the equation of heat conduction, factors that determine the rate of heat conduction on metals are conductivity of metal (k), the difference of temperature (T) and the length of metal (l).

6. Two rods of the same size but different type, as shown in the figure below. The thermal conductivity of rod P is 2 times the thermal conductivity of rod Q. What is the temperature between both rods.

Known :

Both rods have the same size.Temperature and heat – problems and solutions 5

Thermal conductivity of rod P (kP) = 2k

The thermal conductivity of rod Q (kQ) = k

Wanted: The temperature between both rods

Solution :

The equation of the heat conduction :

Temperature and heat – problems and solutions 6

Q/t = the rate of heat conduction, k = thermal conductivity, A = the cross-sectional area, T1-T2 = the change in temperature, l = the length of the rod.

The temperature at the center between both rods :

Temperature and heat – problems and solutions 7

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Black principle

8. 100-gram oil at 20oC and 50-gram iron at 75 oC are placed in 200-gram iron container. The increase in temperature of the container is 5oC and the specific heat of oil is 0.43 cal/g oC. What is the specific heat of the iron?

Known :

Mass of iron container (m) = 200 gr

The initial temperature of the iron container (T1) = the temperature of oil = 20oC

The final temperature of the iron container (T2) = 20oC + 5oC = 25oC

Mass of oil (m) = 100 gram

The specific heat of oil (coil) = 0.43 cal/g oC

The initial temperature of oil (T1) = 20oC

The final temperature of oil (T2) = 20oC + 5oC = 25oC

Mass of iron (m) = 50 gram

The initial temperature of oil (T1) = 75oC

The final temperature of oil (T2) = 25oC

Wanted : The specific heat of iron (c iron)

Solution :

Heat released by iron :

Q = m c ΔT = (50)(c)(75-25) = (50)(c)(50) = 2500c calorie

Heat absorbed by the iron container :

Q = m c ΔT = (200)(c)(25-20) = (200)(c)(5) = 1000c calorie

Heat absorbed by oil :

Q = m c ΔT = (100)(0.43)(25-20) = (43)(5) = 215 calorie

Black principle states that in a isolated system, heat released by the hotter object, absorbed by the cooler object.

Q release = Q absorb

2500c = 1000c + 215

2500c – 1000c = 215

1500c = 215

c = 215/1500

c = 0.143 cal/g oC

9. A 200-gram water at 20°C placed in 50-gram ice at -2°C. If the change of heat just between water and ice, what is the final temperature of the mixture? The specific heat of water is 1 cal/gr°C, the specific heat of ice is 0.5 cal/gr°C, the heat of fusion for ice is 80 cal/gr.

Known :

Mass of water (mwater) = 200 gram

Temperature of water (Twater) = 20oC

The specific heat of water (cwater) = 1 cal/gr°C

Mass of ice (mice) = 50 gram

Temperature of ice (Tice) = -2oC

The specific heat of ice (cice) = 0.5 cal/gr°C

The heat of fusion for ice (L) = 80 cal/gr

Solution :

Heat to increases ice from -2oC to 0oC :

Q = m c ΔT

Q = (50 gram)(0.5 cal/gr°C)(0oC – (-2oC))

Q = (50)(0.5 cal)(2)

Q = 50 calorie

Heat for melting all ice :

Q = m L = (50 gram)(80 cal/gram) = 4000 calorie

Heat for decrease temperature of all water from 20oC to 0oC :

Q = m c ΔT

Q = (200 gram)(1 cal/gr°C)(0oC – (20oC))

Q = (200)(1 cal)(-20)

Q = -4000 calorie

Plus sign indicates that the heat is added, the minus sign indicates that heat released.

50-calorie of heat needed to increase the temperature of ice to 0oC and 4000-calorie needed to melting all ice. Total heat = 4050 calorie. The heat released by water is 4000 calorie.

Some of the ice not melting, so the final temperature of ice and water is 0oC.

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10. A 200-gram aluminum at 20oC placed in 100-gram water at 80oC in a container. The specific heat of aluminum is 0.22 cal/g oC and the specific heat of water is 1 cal/g oC. What is the final temperature of aluminum?

Known :

Mass of aluminum = 200 gram

Temperature of aluminum = 20oC

Mass of water = 100 gram

Temperature of water = 80oC

The specific heat of aluminum = 0.22 cal/g oC

The specific heat of water = 1 cal/g oC

Wanted: The final temperature of aluminum

Solution :

Aluminum and water in thermal equilibrium so that the final temperature of aluminum = the final temperature of water.

Heat released by hot water (Q release) = heat absorbed by aluminum (Q absorb)

mwater c (ΔT) = maluminum c (ΔT)

(100)(1)(80 – T) = (200)(0.22)(T – 20)

(100)(80 – T) = (44)(T – 20)

8000 – 100T = 44T – 880

8000 + 880 = 44T + 100T

8880 = 144T

T = 62oC

11. A 50-gram metal at 85 °C placed in 50 gram water at 29.8 °C. The specific heat of water = 1 cal.g —1 .°C—1. The final temperature is 37 °C. What is the specific heat of metal.

Known :

Mass of metal (mmetal) = 50 gram

Temperature of metal = 85oC

Mass of water (mwater) = 50 gram

Temperature of water = 29,8oC

The specific heat of water (cwater) = 1 cal.g -1 .°C-1

The final temperature of water = 37oC

Wanted : The specific heat of metal (c metal)

Solution :

Heat released by hot metal (Q release) = heat absorbed by water (Q absorb)

mmetal c (ΔT) = mwater c (ΔT)

(50)(c)(85 – 37) = (50)(1)(37 – 29.8)

(c)(85 – 37) = (1)(37 – 29.8)

48 c = 7.2

c = 0.15 cal.g -1 .°C-1

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12. A block of ice with mass of 50-gram at 0°C and 200-gram water at 30°C, placed in a container. . If the specific heat of water is 1 cal.g– 1 °C –1 and the heat of fusion for ice is 80 cal.g 1. What is the final temperature of the mixture.

Known :

Mass of ice (mice) = 50 gram

The temperature of ice = 0°C

Mass of water (mwater) = 200 gram

Temperature of water = 30oC

The specific heat of water (cwater) = 1 cal.g– 1 °C –1

The heat of fusion for ice (Lice) = 80 cal.g –1

Wanted: The final temperature

Solution :

Estimate the final condition :

Heat released by water to decrease its temperature from 30oC to 0oC :

Qrelease = mwater cwater (ΔT) = (200)(1)(30-0) = (200)(30) = 6000

Heat needed to melting all ice :

Q = mice Lice= (50)(80) = 4000

Heat used to melting all ice is 4000, while heat released by water is 6000. Can be concluded that the final temperature of the mixture above 0oC.

Black principle :

Heat released by water = heat for melting all ice + heat to increase the temperature of ice.

(mwater)(cwater)(ΔT) = (mice)(Lice) + (mice)(cwater)(ΔT)

(200)(1)(30-T) = (50)(80) + (50)(1)(T-0)

(200)(30-T) = (50)(80) + (50)(T-0)

6000 – 200T = 4000 + 50T – 0

6000 – 4000 = 50T + 200T

2000 = 250T

T = 2000/250

T = 8oC