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Question : A 10-cm-diameter copper ball is to be heated from 100°C to a...

A 10-\mathrm{cm} -diameter copper ball is to be heated from 100^{\circ} \mathrm{C} to an average temperature of 150^{\circ} \mathrm{C} in 30 minutes (Fig. 1-14 ). Taking the average density and specific heat of copper in this temperature range to be \rho=8950 \mathrm{~kg} / \mathrm{m}^{3}and c_{p}=0.395 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, respectively, determine (a) the total amount of heat transfer to the copper ball, (b) the average rate of heat transfer to the ball, and (c) the average heat f \| x.

 

 

 

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The copper ball is to be heated from 100^{\circ} \mathrm{C} to 150^{\circ} \mathrm{C}. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined. Assumptions Constant properties can be used for copper at the average temperature. Properties The average density and specific heat of copper are given to be \rho=8950 \mathrm{~kg} / \mathrm{m}^{3}and c_{p}=0.395 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}

Analysis (a) The amount of heat transferred to the copper ball is simply the change in its internal energy, and is determined from Energy transfer to the system = Energy increase of the system

Q=\Delta U=m c_{\mathrm{avg}}\left(T_{2}-T_{1}\right)

where m=\rho V=\frac{\pi}{6} \rho D^{3}=\frac{\pi}{6}\left(8950 \mathrm{~kg} / \mathrm{m}^{3}\right)(0.1 \mathrm{~m})^{3}=4.686 \mathrm{~kg}

Substituting, Q=(4.686 \mathrm{~kg})\left(0.395 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(150-100){ }^{\circ} \mathrm{C}=92.6 \mathrm{~kJ}

Therefore, 92.6 \mathrm{~kJ} of heat needs to be transferred to the copper ball to heat it from 100^{\circ} \mathrm{C} to 150^{\circ} \mathrm{C}.

(b) The rate of heat transfer normally changes during a process with time. However, we can determine the average rate of heat transfer by dividing the total amount of heat transfer by the time interval. Therefore,

\dot{Q}_{\text {avg }}=\frac{Q}{\Delta t}=\frac{92.6 \mathrm{~kJ}}{1800 \mathrm{~s}}=0.0514 \mathrm{~kJ} / \mathrm{s}=51.4 \mathrm{~W}

(c) Heat flux is defined as the heat transfer per unit time per unit area, of the rate of heat transfer per unit area. Therefore, the average heat flux in this case is

\dot{q}_{\mathrm{avg}}=\frac{\dot{Q}_{\text {avg }}}{A}=\frac{\dot{Q}_{\text {avg }}}{\pi D^{2}}=\frac{51.4 \mathrm{~W}}{\pi(0.1 \mathrm{~m})^{2}}=1636 \mathrm{~W} / \mathrm{m}^{2}

Dscussion Note that heat flux may vary with location on a surface. The value calculated above is the average heat flux over the entire surface of the ball.