Eggs with a mass of \(0.15 \mathrm{~kg}\) per egg and a specific heat of \(3.32 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\) are cooled from \(32^{\circ} \mathrm{C}\) to \(10^{\circ} \mathrm{C}\) at a rate of 200 eggs per minute. The rate of heat removal from the eggs is (a) \(7.3 \mathrm{~kW}\) (b) \(53 \mathrm{~kW}\) (c) \(17 \mathrm{~kW}\) (d) \(438 \mathrm{~kW}\) (e) \(37 \mathrm{~kW}\)

Short Answer

Expert verified
Question: The eggs have a mass of 0.15 kg and specific heat of 3.32 kJ/kg°C. They are cooled down from 32°C to 10°C at a rate of 200 eggs per minute. Calculate the rate of heat removal from the eggs. Answer: The rate of heat removal from the eggs is approximately 36.38 kW.

Step by step solution

01

Calculate the heat removed from one egg

Calculate the heat removed from one egg using the formula: Q = mcΔT where Q is the heat removed, m is the mass of the egg (0.15 kg), c is the specific heat capacity (3.32 kJ/kg°C), and ΔT is the change in temperature (from 32°C to 10°C). ΔT = 32 - 10 = 22°C Q = (0.15 kg) × (3.32 kJ/kg°C) × (22°C) = 10.914 kJ
02

Convert the heat removed into watts

We need to convert the heat removed per egg into watts. 1 kJ = 1000 J, and since there are 60 seconds in a minute, we need to divide the value by 60 to get the heat removed per second: Q = (10.914 kJ) × (1000 J/kJ) × (1/60) = 181.9 J/s = 181.9 W
03

Calculate the heat removed for 200 eggs per minute

Now, we need to calculate the heat removed for all the eggs cooled per minute. Since there are 200 eggs cooled per minute, we can simply multiply the heat removed per egg with the number of eggs: Total heat removed = (181.9 W) × (200 eggs) = 36380 W
04

Convert the heat removed into kilowatts

Finally, we need to convert the total heat removed per minute into kilowatts: Total heat removed = 36380 W = 36.38 kW The closest answer in the given options is (e) 37 kW.

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Most popular questions from this chapter

A 10 -cm-high and 20-cm-wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of \(0.12 \mathrm{~W}\) and transferring it by convection and radiation to the surrounding medium at \(40^{\circ} \mathrm{C}\). Heat transfer from the back surface of the board is negligible. If the combined convection and radiation heat transfer coefficient on the surface of the board is \(22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the average surface temperature of the chips is (a) \(41^{\circ} \mathrm{C}\) (b) \(54^{\circ} \mathrm{C}\) (c) \(67^{\circ} \mathrm{C}\) (d) \(76^{\circ} \mathrm{C}\) (e) \(82^{\circ} \mathrm{C}\)

The deep human body temperature of a healthy person remains constant at \(37^{\circ} \mathrm{C}\) while the temperature and the humidity of the environment change with time. Discuss the heat transfer mechanisms between the human body and the environment both in summer and winter, and explain how a person can keep cooler in summer and warmer in winter.

The inner and outer surfaces of a 4-m \(\times 7-\mathrm{m}\) brick wall of thickness \(30 \mathrm{~cm}\) and thermal conductivity \(0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) are maintained at temperatures of \(26^{\circ} \mathrm{C}\) and \(8^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer through the wall, in W.

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