Thermodynamics and Heat Transfer: Sample Numerical Questions with Solutions 

These problems cover various core concepts—conduction, convection, radiation, and basic thermodynamic laws

1. Heat Conduction Through a Metal Rod

Question
A cylindrical metal rod is 0.5 m long, with a cross-sectional area of

$1.5 \times 10^{-3}\,\mathrm{m^2}$

The thermal conductivity k of the metal is 200W/(m.K). One end of the rod is maintained at 100ºC and the other end at 20ºC .

Calculate the steady-state heat flow rate (Q) through the rod.

Solution

1. Formula for steady-state conduction:
   $$\dot{Q}=\frac{kA(T_{\text{hot}}-T_{\text{cold}})}{L}$$

    

2. Plug in the values:
   $$\dot{Q}=\frac{200\mathrm{W}\mathrm{/}(\mathrm{m}\cdot \mathrm{K})\times 1.5\times {10}^{-3}{\mathrm{m}}^2\times (100-20)\mathrm{K}}{0.5\mathrm{m}}$$

    

3. Numerical calculation:

   $$\dot{Q} = \frac{200 \times 1.5 \times 10^{-3} \times 80}{0.5} = \frac{200 \times 1.5 \times 80 \times 10^{-3}}{0.5} = \frac{200 \times 1.5 \times 80}{0.5} \times 10^{-3}$$ $$= \frac{200 \times 1.5 \times 80}{0.5} \times 10^{-3} = \frac{24{,}000}{0.5} \times 10^{-3} = 48{,}000 \times 10^{-3} = 48\,\mathrm{W}$$
4. Result:
   $$\dot{Q} = 48\,\mathrm{W}$$

    

2. Convective Heat Transfer From a Flat Plate

Question

Air at

$25^\circ\mathrm{C}$

flows over a flat plate maintained at

$75^\circ\mathrm{C}$

. If the convective heat transfer coefficient

$h$

is

$25\,\mathrm{W/(m^2 \cdot K)}$

and the plate’s surface area is

$0.2\,\mathrm{m^2}$

, calculate the rate of heat transfer by convection.

Solution

1. Formula for convective heat transfer:
   $$\dot{Q} \;=\; h \, A \,(T_{\text{surface}} – T_{\text{fluid}})$$

    

    

2. Plug in the values (converting temperatures to Kelvin differences is not strictly necessary here, as it’s a temperature difference):
   $$\dot{Q} \;=\; 25 \,\mathrm{W/(m^2 \cdot K)} \;\times\; 0.2\,\mathrm{m^2} \;\times\; (75 – 25)\,\mathrm{K}$$

    

    

3. Numerical calculation:
   $$\dot{Q} = 25 \times 0.2 \times 50 = 25 \times 10 = 250\,\mathrm{W}$$

    

    

4. Result:
   $$\dot{Q} = 250\,\mathrm{W}$$

    

    

3. Radiation Heat Transfer Between Two Surfaces

Question
Two large parallel plates each have an emissivity of

$0.8$

. Plate A is at

$500\,\mathrm{K}$

and Plate B is at

$300\,\mathrm{K}$

. Calculate the net radiative heat transfer rate per unit area (

$q_{\text{net}}$

) from Plate A to Plate B. Use

$\sigma = 5.67 \times 10^{-8}\,\mathrm{W/(m^2 \cdot K^4)}$

.

Solution

1. Formula for net radiation exchange between two large parallel plates:
   $$q_{\text{net}} \;=\; \frac{\sigma (T_A^4 – T_B^4)}{\frac{1}{\varepsilon_A} + \frac{1}{\varepsilon_B} – 1}$$

    

   Here,

   $\varepsilon_A = \varepsilon_B = 0.8$

   εA=εB=0.8.

2. Calculate the denominator for emissivities:
   $$\frac{1}{0.8} + \frac{1}{0.8} – 1 = 1.25 + 1.25 – 1 = 2.5 – 1 = 1.5$$

    

    

3. Calculate
   $T_A^4 – T_B^4$

    

   TA4−TB4:

   $$T_A^4 = (500)^4 = 500^4 = 6.25 \times 10^{10}\,\mathrm{K^4}$$

    

   $$T_B^4 = (300)^4 = 300^4 = 8.1 \times 10^{9}\,\mathrm{K^4}$$

    

   $$T_A^4 – T_B^4 = 6.25 \times 10^{10} – 8.1 \times 10^9 = 6.25 \times 10^{10} – 0.81 \times 10^{10} = (6.25 – 0.81) \times 10^{10} = 5.44 \times 10^{10}$$

    

4. Compute
   $q_{\text{net}}$

    

   qnet:

   $$q_{\text{net}} = \frac{5.67 \times 10^{-8} \,\times\, 5.44 \times 10^{10}}{1.5} = \frac{5.67 \times 5.44 \times 10^{2}}{1.5}$$ $$= \frac{30.85 \times 10^2}{1.5} = \frac{3085}{1.5} \approx 2057 \,\mathrm{W/m^2}$$

    

5. Result:
   $$q_{\text{net}} \approx 2057\,\mathrm{W/m^2}$$

    

    

(Note: Depending on rounding, results may slightly differ.)

4. First Law of Thermodynamics in a Piston-Cylinder

Question
A piston-cylinder device contains

$0.1\,\mathrm{kg}$

of a gas. The gas is heated from

$300\,\mathrm{K}$

 to

$400\,\mathrm{K}$

 at constant pressure. If the specific heat at constant pressure

$c_p = 1.0\,\mathrm{kJ/(kg \cdot K)}$

, calculate the heat added (

$Q$

) in kilojoules, assuming negligible work other than boundary work at constant pressure.

Solution

1. Formula for heat added at constant pressure:
   $$Q=m{c}_p\mathrm{\Delta }T$$

    

2. Plug in the values:
   $$\Delta T = (400 – 300)\,\mathrm{K} = 100\,\mathrm{K}$$

    

   $$Q = 0.1\,\mathrm{kg} \,\times\, 1.0\,\mathrm{kJ/(kg \cdot K)} \,\times\, 100\,\mathrm{K}$$
3. Numerical calculation:
   $$Q = 0.1 \times 1.0 \times 100 = 10\,\mathrm{kJ}$$

    

4. Result:
   $$Q = 10\,\mathrm{kJ}$$

    

5. Efficiency of a Carnot Engine

Question
A Carnot heat engine operates between two reservoirs at

$T_H = 500\,\mathrm{K}$

and

$T_C = 300\,\mathrm{K}$

. Calculate the maximum theoretical efficiency (

$\eta_{\text{Carnot}}$

) of this engine as a percentage.

Solution

1. Carnot efficiency formula:
   $$\eta_{\text{Carnot}} \;=\; 1 \;-\; \frac{T_C}{T_H}$$

    

2. Plug in the temperatures:
   $$\eta_{\text{Carnot}} = 1 – \frac{300}{500} = 1 – 0.6 = 0.4$$

    

3. Convert to a percentage:
   $$0.4 \times 100\% = 40\%$$

    

4. Result:
   $$\eta_{\text{Carnot}} = 40\%$$

    

6. Mass Flow Rate in a Steady-Flow System

Question
In a steady-flow device, water enters at a mass flow rate of

$0.02\,\mathrm{kg/s}$

. It gains

$20\,\mathrm{kJ/kg}$

of energy from a heat source as it passes through. If there is no significant change in kinetic or potential energy, how much heat transfer (

$\dot{Q}$

˙) occurs per second?

Solution

1. 1. First Law for steady-flow with negligible changes in KE and PE:
      $$\dot{Q} \;=\; \dot{m} \,\Delta h \quad (\text{if } w \approx 0)$$

       

      $\Delta h$ is the change in specific enthalpy, which equals

      $20\,\mathrm{kJ/kg}$

       in the form of heat gained.

   2. Plug in the values:
      $$\dot{Q} = 0.02\,\mathrm{kg/s} \,\times\, 20\,\mathrm{kJ/kg}$$

       

   3. Calculate in
      $\mathrm{kJ/s}$

       

      , and the temperature difference across the entire wall is

      $20\,\mathrm{K}$

      . Calculate the heat flux (

      $\dot{Q}/A$

      ) through the wall.

Solution

1. 1. 1. Equivalent thermal resistance
         $R_{\text{total}}$

          

         Rtotal in series:

         $$R_{\text{total}} \;=\; \frac{L_1}{k_1} \;+\; \frac{L_2}{k_2}$$
      2. Compute each term:
         $$R_1 = \frac{0.02}{0.8} = 0.025\,\mathrm{m^2 K/W}$$

          

          

         $$R_2 = \frac{0.01}{0.04} = 0.25\,\mathrm{m^2 K/W}$$

          

         $$R_{\text{total}} = 0.025 + 0.25 = 0.275\,\mathrm{m^2 K/W}$$

          

      3. Overall heat transfer:
         $$\dot{Q} \;=\; \frac{\Delta T}{R_{\text{total}}} \times A$$

          

          

      4. Heat flux (per unit area) is:
         $$\frac{\dot{Q}}{A} = \frac{\Delta T}{R_{\text{total}}} = \frac{20}{0.275} \approx 72.73\,\mathrm{W/m^2}$$

          

          

      5. Result:
         $$\text{Heat flux} = 72.73\,\mathrm{W/m^2}$$

          

          

$$\text{Heat flux} = 72.73\,\mathrm{W/m^2}$$

8. Determining Outlet Temperature in a Heat Exchanger

Question
Water enters a heat exchanger at

$15^\circ\mathrm{C}$

and flows at

$0.05\,\mathrm{kg/s}$

. It absorbs heat at a rate of

$10\,\mathrm{kW}$

10kW. Given the specific heat of water

$c_p \approx 4.2\,\mathrm{kJ/(kg \cdot K)}$

cp, determine the water outlet temperature.

Solution

1. 1. 1. First, relate heat absorbed (
         $\dot{Q}$

          

         ) to mass flow rate (

         $\dot{m}$

         ), specific heat (

         $c_p$

         ), and temperature change (

         $\Delta T$

         ):

         $$\dot{Q} = \dot{m} \, c_p \, \Delta T$$

          

      2.  

         :

         $$\Delta T = \frac{\dot{Q}}{\dot{m} \, c_p}$$
      3. Plug in the values (converting kW to kJ/s if needed—1 kW = 1 kJ/s):
         $$\dot{Q} = 10\,\mathrm{kW} = 10\,\mathrm{kJ/s}$$

          

         $$\dot{m} = 0.05\,\mathrm{kg/s}, \quad c_p = 4.2\,\mathrm{kJ/(kg \cdot K)}$$ $$\Delta T = \frac{10\,\mathrm{kJ/s}}{0.05\,\mathrm{kg/s} \times 4.2\,\mathrm{kJ/(kg \cdot K)}}$$ $$= \frac{10}{0.21} = 47.62\,\mathrm{K} \approx 47.6^\circ\mathrm{C}$$
      4. Outlet temperature:
         $$T_{\text{out}} = T_{\text{in}} + \Delta T = 15^\circ\mathrm{C} + 47.6^\circ\mathrm{C} = 62.6^\circ\mathrm{C}$$

          

      5. Result:
         $$T_{\text{out}} \approx 62.6^\circ\mathrm{C}$$

          

9. Calculating Entropy Change in an Isothermal Process

Question

$0.1\mathrm{kg}$ of an ideal gas undergoes an isothermal expansion at

$T = 300\,\mathrm{K}$

from a volume of

$0.02\,\mathrm{m^3}$

to

$0.04\,\mathrm{m^3}$

3. The gas constant

$R$

R for the gas is

$0.287\,\mathrm{kJ/(kg \cdot K)}$

. Calculate the entropy change

$\Delta S$

 of the gas in

$\mathrm{kJ/(kg \cdot K)}$

.

Solution

1. 1. 1. Entropy change for an isothermal process (ideal gas):
         $$\Delta S = R \,\ln\!\biggl(\frac{V_2}{V_1}\biggr)$$

          

          This is per unit mass for a gas with specific gas constant R.

         $$
      2. Plug in volumes and R
         $$\Delta S = 0.287\,\mathrm{kJ/(kg \cdot K)} \times \ln\!\biggl(\frac{0.04}{0.02}\biggr)$$

          

      3. Calculate the logarithmic term:
         $$\ln\!\biggl(\frac{0.04}{0.02}\biggr) = \ln(2) \approx 0.693$$

          

      4. Numerical result:
         $$\Delta S = 0.287 \times 0.693 \approx 0.199\,\mathrm{kJ/(kg \cdot K)}$$

          

      5. Result:
         $$\Delta S \approx 0.20\,\mathrm{kJ/(kg \cdot K)}$$

          

10. Work Done in an Adiabatic Expansion

Question
An ideal gas expands adiabatically from an initial state of

$P_1 = 5\,\mathrm{bar}, V_1 = 0.01\,\mathrm{m^3}$

to a final volume of

$V_2 = 0.02\,\mathrm{m^3}$

. The process follows

$PV^\gamma = \text{constant}$

, with

$\gamma = 1.4$

. If the gas is assumed ideal, calculate the work done by the gas (

$W$

) in joules. Assume no heat transfer (

$Q = 0$

).

Solution

1. 1. 1. Relationship for an adiabatic process (ideal gas):
         $$P_1 V_1^\gamma = P_2 V_2^\gamma$$

          

         We need to find

         $P_2$

         first, then compute the work.

      2. Find
         $P_2$

          

         :

         $$P_2 = P_1 \biggl(\frac{V_1}{V_2}\biggr)^\gamma$$

         Converting

         $P_1$

         into Pa (SI units):

         $$P_2 = 5 \times 10^5 \,\mathrm{Pa} \times \biggl(\frac{0.01}{0.02}\biggr)^{1.4} = 5 \times 10^5 \,\mathrm{Pa} \times \bigl(0.5\bigr)^{1.4}$$