The heat transfer coefficient for air flowing over asphere is to be determined by observing temperature-time history of a sphere fabricated from pure copper. The sphere, which is 12.7 mm in diameter, is at 66oC before it is inserted into an airstream having a temperature of 27oC. A thermocouple on the outer surface of the sphere indicates 55oC 69 s after the sphere is inserted. (a) Assume and then justify that the sphere behaves as spacewise isothermal object. (b) Calculate the heat transfer coefficient, h. (c) Calculate the amount of heat lost by the sphere during this time, i.e., 69 s.

(a) The assumption that the sphere behaves as a spacewise isothermal object implies that the temperature is uniform across the sphere's volume at any given time. This assumption is justified for a sphere made of pure copper because copper has a high thermal conductivity, meaning it can rapidly conduct heat throughout its volume. Therefore, when exposed to a different temperature environment, the temperature within the sphere would equalize quickly, allowing us to treat the sphere as if it has a uniform temperature.

(b) To calculate the heat transfer coefficient, h, we can use Newton's law of cooling, which states that the rate of heat transfer (Q/t) from the sphere to the surrounding air is proportional to the difference in temperature between the sphere and the air, and the surface area (A) through which the heat is being transferred. The equation can be written as:

Q/t = h * A * (Tsphere - Tair)

Where: - Q is the amount of heat transferred, - t is the time, - h is the heat transfer coefficient, - A is the surface area of the sphere, - Tsphere is the temperature of the sphere, - Tair is the temperature of the air.

Given that we do not have the value of Q directly, we should work with the rate of temperature change. Assuming that the sphere is spacewise isothermal, we can write the heat transfer as a function of temperature change in the sphere over time:

m * cp * (dT/dt) = h * A * (Tsphere - Tair)

Where: - m is the mass of the sphere, - cp is the specific heat capacity of copper, - dT/dt is the rate of temperature change in the sphere.

We need to rearrange the equation and solve for h:

h = m * cp * (dT/dt) / [A * (Tsphere - Tair)]

First, we'll need to calculate the mass of the sphere (m = volume * density) and the surface area (A = 4πr²).

The volume of the sphere, V = (4/3)πr³, where r is the radius.

r = 12.7 mm / 2 = 6.35 mm = 6.35 * 10^-3 m (converted to meters) Density of copper, ρ (rho) = approximately 8960 kg/m³. Specific heat capacity of copper, cp = approximately 385 J/kg·K.

So,

m = V * ρ = (4/3)π(6.35 * 10^-3 m)³ * 8960 kg/m³

Now, find A,

A = 4 * π * r² = 4 * π * (6.35 * 10^-3 m)²

And,

(dT/dt) ≈ (Tsphere_initial - Tsphere_final) / t = (66 - 55) °C / 69 s = 11 °C / 69 s

Finally, substitute these values (where necessary conversion to consistent units is needed) and solve for "h".

Remember, Tsphere is the initial temperature of the sphere, Tair is the temperature of the airstream, and t is the time in seconds since insertion.

(c) To calculate the amount of heat lost by the sphere, we use the heat transfer equation, Q = m * cp * ΔT.

ΔT is the change in temperature, which is Tsphere_initial - Tsphere_final = 66 - 55 = 11°C (which must be converted to Kelvin by adding 273.15 if needed, although for a difference in temperature this conversion is unnecessary as 1°C equates to 1 Kelvin difference).

Plug in the values for the mass m and the specific heat capacity cp (converted into appropriate units if needed) of the sphere and ΔT into the formula to get Q, the amount of heat lost over the period of 69 seconds.

Please note that this explanation sets up the necessary framework for solving the problem and assumes that all quantities are measured in coherent SI units. In practice, additional intermediate calculations may be needed to ensure that units are consistent before substituting into the formulas.