The free convection heat transfer coefficient on a thin 
hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is 
isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at 
the instant of time when the plate temperature is 245°C 
and the change in plate temperature with time (dT/dt) 
is −0.028 K/s. The ambient air temperature is 25°C and 
the plate measures 0.4 × 0.4 m with a mass of 4.25 kg 
and a specific heat of 2770 J/kg ⋅K

To determine the free convection heat transfer coefficient \( h \) at the given instant, we can use the concept of energy balance on the thin plate. The heat loss by convection is related to the rate of temperature change in the plate using the following principles.

### Governing Equation:
The rate of energy loss due to convection from the plate is balanced by the rate of internal energy change of the plate:

\[
q_{\text{conv}} = m c_p \frac{dT}{dt}
\]

where:
- \( q_{\text{conv}} \) is the rate of heat loss by convection,
- \( m \) is the mass of the plate,
- \( c_p \) is the specific heat of the plate,
- \( \frac{dT}{dt} \) is the rate of change of the plate's temperature with respect to time.

The convection heat loss is also given by:

\[
q_{\text{conv}} = h A (T_{\text{plate}} - T_{\infty})
\]

where:
- \( h \) is the convection heat transfer coefficient (what we're solving for),
- \( A \) is the surface area of the plate,
- \( T_{\text{plate}} \) is the temperature of the plate,
- \( T_{\infty} \) is the ambient air temperature.

### Step-by-Step Solution:

1. **Calculate the surface area of the plate**:
   The plate is a square with dimensions \( 0.4 \, \text{m} \times 0.4 \, \text{m} \), so the surface area is:
   \[
   A = 0.4 \times 0.4 = 0.16 \, \text{m}^2
   \]

2. **Substitute the known values into the energy balance equation**:
   The rate of heat loss by convection is equal to the rate of temperature change, which can be written as:
   \[
   h A (T_{\text{plate}} - T_{\infty}) = m c_p \frac{dT}{dt}
   \]

   We are given:
   - \( T_{\text{plate}} = 245^\circ \text{C} = 245 + 273 = 518 \, \text{K} \),
   - \( T_{\infty} = 25^\circ \text{C} = 25 + 273 = 298 \, \text{K} \),
   - \( m = 4.25 \, \text{kg} \),
   - \( c_p = 2770 \, \text{J/kg⋅K} \),
   - \( \frac{dT}{dt} = -0.028 \, \text{K/s} \),
   - \( A = 0.16 \, \text{m}^2 \).

3. **Rearrange the equation to solve for \( h \)**:
   \[
   h = \frac{m c_p \frac{dT}{dt}}{A (T_{\text{plate}} - T_{\infty})}
   \]

4. **Substitute the values**:
   \[
   h = \frac{4.25 \times 2770 \times (-0.028)}{0.16 \times (518 - 298)}
   \]

   Simplifying:

   \[
   h = \frac{4.25 \times 2770 \times (-0.028)}{0.16 \times 220}
   \]

   \[
   h = \frac{-329.86}{35.2}
   \]

   \[
   h \approx -9.37 \, \text{W/m}^2 \cdot \text{K}
   \]

Since the heat transfer coefficient is positive by definition, we take the absolute value:

\[
h \approx 9.37 \, \text{W/m}^2 \cdot \text{K}
\]

### Conclusion:
The free convection heat transfer coefficient at the given conditions is approximately \( 9.37 \, \text{W/m}^2 \cdot \text{K} \).

Would you like further details or clarifications? 

Here are 5 related questions to extend this problem:

1. How does the size of the plate affect the convection heat transfer coefficient?
2. What would be the effect of including radiation in the heat transfer calculation?
3. How would a change in the ambient air temperature affect the heat transfer coefficient?
4. Could this analysis be applied to a non-isothermal plate? What adjustments would be needed?
5. How does the specific heat capacity of the material influence the cooling rate?

**Tip**: When solving for heat transfer coefficients, ensure that the physical properties of the material (such as specific heat and surface area) are well-defined for accuracy.

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is 245°C and the change in plate temperature with time (dT/dt) is −0.028 K/s. The ambient air temperature is 25°C and the plate measures 0.4 × 0.4 m with a mass of 4.25 kg and a specific heat of 2770 J/kg ⋅K.

Determine the free convection heat transfer coefficient for a thin hot vertical plate suspended in still air. The problem provides a detailed step-by-step solution using thermodynamic principles and heat transfer equations. The plate is isothermal, and radiation is neglected. Key data includes a plate temperature of 245°C, mass of 4.25 kg, specific heat of 2770 J/kg⋅K, and dT/dt of −0.028 K/s.

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