Physics 9-Ch8 (Thermal Properties of Matter)-Numerical Problems

 

Table of Contents

• what is its value on the Fahrenheit scale
• Normal human body temperature is 98.6°F, convert it into Celsius scale and Kelvin scale.
• Calculate the increase in the length of an aluminum bar 2 m
• The thermal coefficient of volume expansion of air
• How much heat is required to increase the temperature
• An electric heater supplies heat at the rate of 1000J per second.
• How much ice will melt by 5000 J of heat
• Find the quantity of heat needed to melt 100g of ice
• How much heat is required to change 100g of water at 100°C into steam
• Find the temperature of the water after passing 5 g of steam

The temperature of water in a beaker is 50°C, what is its value on the Fahrenheit scale? (122°F)

Difficulty: Easy

 Solution:   Temperature in Celsius scale = C = 50°C

                    The temperature on the Fahrenheit scale = F =?

                    F = 1.8C + 32

                    F = $1.8\times$ 50 + 32 = 90 + 32

                    F = 122°F

Normal human body temperature is 98.6°F, convert it into Celsius scale and Kelvin scale.                                                                                                            (37°C,310K)

Difficulty: Easy

Solution:   

The temperature on Fahrenheit scale = 98.6°F

The temperature on the Celsius scale =?

The temperature on the Kelvin scale =?

 

F = 1.8C + 32

1.8C = F – 32

1.8C = 98.6 -32

1.8C = 66.6

C =$\frac{(66.6)}{(1.8)}$ = 37°C

 

T(K) = C + 273

T(K) = 37 + 273

T(K) = 310K

Calculate the increase in the length of an aluminum bar 2 m long when heated from 0°C to 20°C, if the thermal coefficient of linear expansion of aluminum is $2.5\times \text{〖}10{\text{〗}}^{-5}{�}^{-1}$. (0.1cm)

Difficulty: Easy

Solution:       Original length of rod = ${�}_{(0=)}$ 2m
                  Initial temperature = ${�}_{(0=)}$  0°C = 0 + 273 = 273 K
              Final temperature = T =  20°C = 20 + 273 = 293 K
           Change in temperature = $\mathrm{△}�$   = $�-{�}_0$ = 293 – 273 = 20 K
         Coefficient of linear expansion of aluminum = α = $2.5\times 10^{-5}{�}^{-1}$
             Increase in volume $\mathrm{△}�$ = ?
                                      $\mathrm{△}�=�{�}_0\mathrm{△}$ T   
                                        $\mathrm{△}�=2.5\times 10^{-5}\times 20$
                                       $\mathrm{△}�=100\times 10^{-5}$
                $\mathrm{△}�=0.001�=0.001\times$ 100 = 0.1cm

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A balloon contains $1.2{�}^3$ air at 15°C. Find its volume at 40°C. The thermal coefficient of volume expansion of air is $3.67\times \text{〖}10{\text{〗}}^{-3}{�}^{-1}$. $(1.3{�}^{(3)}$

Difficulty: Easy

Solution:       Original volume = ${�}_{(0=)}1.2{�}^3$
                      Initial temperature = ${�}_0$ = 15°C = 15 + 273 = 288 K
                       Final temperature = T = 40°C = 40 + 273 = 313 K
                        Change in temperature = $\mathrm{△}�$ = T - ${�}_0$ = 313 – 288 = 25 K
                        Coefficient of volume expansion of air $�$ = $3.67\times 10^{-3}{�}^{-1}$ 
                        Volume = V = ?
                        V = ${�}_0(1+�\mathrm{△}�)$
                        V = $1.2(1+3.67\times 10^{-3}\times 25)$= $1.2(1+91.75\times 10^{-3})$
                            = 1.2(1 + 0.09175) = $1.2\times 1.09175$
                        V = $1.3{�}^3$
                                                       

How much heat is required to increase the temperature of 0.5 kg of water from 10°C to 65°C?                                                                                                                   (115500 J)

Difficulty: Easy

Solution:               Mass of water = m = 0.5 kg
                               Initial temperature = ${�}_1$ = 10°C = 10 + 273 = 283 K
                               Final temperature = ${�}_2$ = 65°C = 65 + 273 = 338 K 
                               Change in temperature = $\mathrm{△}$ T = ${�}_2-{�}_1$ = 338 – 283 = 55 K
                               Heat =$\mathrm{△}�$ = ?
                               $\mathrm{△}�$ = $��\mathrm{△}�$
                               $\mathrm{△}�=0.5\times 2400\times 55$
                               $\mathrm{△}�$ = 115500J

An electric heater supplies heat at the rate of 1000J per second. How much time is required to raise the temperature of 200 g of water from 20°C to 90°C?             (58.8 s)

Difficulty: Medium

Solution:     Power = P = 1000 $�{�}^{-1}$
                    Mass of water = m = 200 g =$\frac{200}{1000}$  = 0.2 kg
                     Initial temperature = ${�}_2$= 20°C = 20 + 273 = 293 K
                    Final temperature =  ${�}_1$= 90°C = 90 + 273 = 363 K
                   Change in temperature = $\mathrm{△}�$ =  ${�}_2-{�}_1$ = 363 – 293 = 70 K
                   Specific heat of water = c = $4200�\text{〖}��{\text{〗}}^{-1}{�}^{-1}$
                 Time = t = ?
                 P =$\frac{�}{�}$ 
Or            P =$\frac{�}{�}$  
Or            P $\times �$ = Q
Or            P $\times �$ = $��\mathrm{△}�$
Or              t=$\frac{(��\mathrm{△}�)}{�}$ 
                   t = $\frac{(0.2\times 4200\times 70)}{1000}$= 58.8 s

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How much ice will melt by 5000 J of heat? Latent heat of fusion of ice = $336000�\text{〖}��{\text{〗}}^{-1}$.                                                                                                                       (149 g )

 

                  

 

Difficulty: Easy

Solution:     Amount of heat required to melt ice = 50000J
                   Latent heat of fusion of ice =  = $336000�\text{〖}��{\text{〗}}^{-1}$
                   Amount of ice = m =?
                       $\mathrm{△}{�}_{�}$=  $�{�}_{�}$
      Or               m =  $\frac{(\mathrm{△}{�}_{�})}{({�}_{�})}$
                         m =$\frac{50000}{336000}$ = 0.1488 kg
                            = $0.1488\times 1000$ =$\frac{1488}{1000}\times$1000 = $148.8�\approx 149�$ 

Find the quantity of heat needed to melt 100g of ice at -10°C into the water at 10°C.    
                                                                                                                                        (39900 J)    
(Note:  Specific heat of ice is $2100�\text{〖}��{\text{〗}}^{-1}{�}^{-1}$, the specific heat of water is $4200�\text{〖}��{\text{〗}}^{-1}{�}^{-1}$, Latent heat of fusion of ice is 336000 J$\text{〖}��{\text{〗}}^{-1})$.

Difficulty: Medium

Solution:   Mass of ice = m = 100 g = $\frac{100}{1000}$  = 0.1 kg

                  Specific heat of ice = c1 = $2100�\text{〖}��{\text{〗}}^{-1}{�}^{-1}$

                   Latent heat of fusion of ice = L = $336000�\text{〖}��{\text{〗}}^{-1}$

                  Specific heat of water = c = $4200�\text{〖}��{\text{〗}}^{-1}{�}^{-1}$

                  Quantity of heat required = Q =?

 

Case I:

            Heat gained by ice from -10°C to 0°C

                                        Q1 = $��\mathrm{△}�$

                                        Q1 = $0.1\times 2100\times 10$ = 2100 J

 

Case II:

            Heat required for ice to melt = Q2 = mL

                                                              = $0.1\times 336000$

                    Q2  = 33600 J

 

Case III:

           The heat is required to raise the temperature of water from 0°C to 10°C

                                                            Q3= $��\mathrm{△}�$

Q3 = $0.1\times 4200\times 10$ = 4200 J 

                                                             Total heat required = Q = Q1 + Q2 + Q3

    Q = 2100 + 33600 + 4200

                     Q = 39900 J

How much heat is required to change 100g of water at 100°C into steam? (Latent heat of vaporization of water is $2.26\times \text{〖}10{\text{〗}}^6�\text{〖}��{\text{〗}}^{-1}$.                $(2.26\times \text{〖}10{\text{〗}}^5�)$

Difficulty: Easy

Solution:       Mass of water = m = 100 g = $\frac{100}{1000}$ = 0.1 kg
              Latent heat of vaporization of water = ${�}_{�}$  = $2.26\times 10^6��{�}^{-1}$
                     Heat required = $\mathrm{△}{�}_{�}$= ?
                                               $\mathrm{△}{�}_{�}=�{�}_{�}$
                                              $\mathrm{△}{�}_{�}$= $0.1\times 2.26\times 10^6$ = $0.226\times 10^6$ = $\frac{226}{1000}\times 10^6$
                                                        = $2.26\times 10^{-1}\times 10^6$  = $2.26\times 10^5$J

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Find the temperature of the water after passing 5 g of steam at 100°C through 500 g of water at 10°C.         (16.2°C)

          (Note: Specific heat of water is $4200�\text{〖}��{\text{〗}}^{-1}{�}^{-1}$, Latent heat of vaporization of water is $2.26\times \text{〖}10{\text{〗}}^6��{�}^{-1})$.

Difficulty: Hard

Solution:    Mass of stream = ${�}_1=5�$ =$\frac{5}{1000}$ kg = 0.005 kg

                   Temperature of stream = ${�}_1$= 100°C

                  Mass of water = ${�}_2$ = 0.5 kg

                  Temperature of water = ${�}_2$= 10°C

                  Final temperature = ${�}_3$= ?

 

Case I:

            Latent heat lost by stream = Q1 = mL

            Q1 = $0.005\times 2.26\times 10^6$ = $11.3\times 10^3$ = 11300 J

 

Case II: 

            Heat lost by stream to attain final temperature Q2 = ${�}_1�\mathrm{△}�$

            Q2 = $0.005\times 4200\times (100-{�}_3)$ 

            Q2 = $21(100-{�}_3)$

 

Case III:

              Heat gained by water Q3 = ${�}_2�\mathrm{△}�$

          Q3 = $0.5\times 4200\times ({�}_3-10)$ 

          Q3 = $2100({�}_3-10)$

                      According to the law of heat exchange.

                       Heat lost by stream = heat gained by water

                                                   Q1 + Q2 = Q3

                              $11300+21(100\text{–}{�}_3)$ = $2100({�}_3-10)$

          $11300+2100-21{�}_3$ = $2100{�}_3$ – 21000

                             $13400+21000\text{–}21{�}_3$ = $2100{�}_3\text{–}21{�}_3$

                                                       34400 = $2121{�}_3$

                                                            ${�}_3$= $\frac{34400}{2121}$ 

                                                            ${�}_3$= 16.2 °C