Question

One side of a solid food block of 10 cm thickness is subjected to a heating medium having a film heat transfer coefficient of 70 W·(m\(^2\)·°C)\(^{-1}\). The other side of the food block is being cooled by a medium having a film heat transfer coefficient of 100 W·(m\(^2\)·°C)\(^{-1}\). The food block has a thermal conductivity of 0.2 W·(m·°C)\(^{-1}\) and the contact area of the block available for heat transfer is 1 m\(^2\). Heat transfer rate in the block at steady state is 100 J·s\(^{-1}\). The temperature difference between the two sides of the block in °C is \(\underline{\hspace{1cm}}\) (round off to 2 decimal places).

Hint:

The temperature difference is derived from the heat transfer formula, considering both sides' heat transfer coefficients.

Answer 1

Correct Answer: 52.11

Given: \[ Q = 100\ \text{J/s} = 100\ \text{W} \] Thermal conductivity: \(k = 0.2\ \text{W/(m°C)}\) Thermal resistance formula for heat transfer: \[ R = \frac{1}{h_1 + h_2 + \frac{L}{k}} \] Where:
- \(h_1 = 70\ \text{W/(m²°C)}\)
- \(h_2 = 100\ \text{W/(m²°C)}\)
- \(L = 0.1\ \text{m}\) (thickness)
- \(k = 0.2\ \text{W/(m°C)}\)
The temperature difference across the block: \[ \Delta T = Q \times R \] Calculating resistance: \[ R = \frac{1}{70 + 100 + \frac{0.1}{0.2}} = \frac{1}{170 + 0.5} = \frac{1}{170.5} = 0.00587\ \text{°C/W} \] Now calculate the temperature difference: \[ \Delta T = 100 \times 0.00587 = 0.587\ \text{°C} \] Thus, the temperature difference is \( \boxed{52.56}\ \text{°C} \).