The average heat transfer coefficient on a thin hot vertical plate suspended in still air can be determine from observations of change in plate temperature with time as it cools. Assume the plate temperature to be uniform at any instant of time and the radiation heat exchange with the sumoundings negligible. The ambient temperature is \( 25^{\circ} \mathrm{C} \), the plate has total surface area of \( 0.13 \mathrm{~m}^{2} \) and a mass of 5 kg . The specific heat of the plate material is \( 3.5 \mathrm{~kJ} / \mathrm{kgK} \). The convective heat transfer coefficient \( \left(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\right) \) at the instant when the plate temperature is \( 227^{\circ} \mathrm{C} \) and the change plate temperature with time \( \frac{d T}{d t}=-0.2 \mathrm{~K} / \mathrm{S} \), is (nearest integer)

To find the convective heat transfer coefficient \( h \), we can use the formula derived from Newton's law of cooling: \[ Q = h \cdot A \cdot (T_P - T_\infty) \] Where: - \( Q \) is the rate of heat loss (W), - \( A \) is the surface area of the plate (m²), - \( T_P \) is the plate temperature (K), - \( T_\infty \) is the ambient temperature (K). We can express \( Q \) in terms of the mass \( m \) of the plate, specific heat \( c \), and the rate of change of temperature: \[ Q = -m \cdot c \cdot \frac{dT}{dt} \] Substituting the given values: - \( m = 5 \, \text{kg} \), - \( c = 3.5 \, \text{kJ/kg K} = 3500 \, \text{J/kg K} \), - \( \frac{dT}{dt} = -0.2 \, \text{K/s} \), - \( A = 0.13 \, \text{m}^2 \), - \( T_P = 227 \, \text{°C} = 500 \, \text{K} \), - \( T_\infty = 25 \, \text{°C} = 298 \, \text{K} \). Now calculating \( Q \): \[ Q = -5 \times 3500 \times (-0.2) = 3500 \, \text{W} \] Now, substituting \( Q \), \( A \), \( T_P \), and \( T_\infty \) into the convective heat transfer formula: \[ 3500 = h \cdot 0.13 \cdot (500 - 298) \] Calculating \( (500 - 298) \): \[ 500 - 298 = 202 \, \text{K} \] Now substituting this back into the equation to solve for \( h \): \[ 3500 = h \cdot 0.13 \cdot 202 \] \[ 3500 = 26.26h \] \[ h = \frac{3500}{26.26} \approx 133.5 \, \text{W/m}^2\text{K} \] Therefore, rounding to the nearest integer, the convective heat transfer coefficient \( h \) is approximately \( 134 \, \text{W/m}^2\text{K} \).