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Bumek [7]
3 years ago
7

4–72 A person puts a few apples into the freezer at 215°C to cool them quickly for guests who are about to arrive. Initially, th

e apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be r 5 840 kg/m3, cp 5 3.81 kJ/kg·K, k 5 0.418 W/m·K, and a 5 1.3 3 1027 m2/s, determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Physics
1 answer:
KiRa [710]3 years ago
4 0

Complete and Clear Question:

A person puts a few apples into the freezer at -15°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be \rho = 840 kg/m3,  c_{p} = 3.81 kJ/kg·K, k = 0.418 W/m·K, and \alpha = 1.3 * 10^{-7} m^{2} /s, determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Answer:

Temperature at the center of the apple, T(t) = 11.215°C

Temperature at the surface of the apple, T(r,t) = 2.68°C

Amount of heat transfer from each apple, Q = 21.47 kJ

Explanation:

For clarity and easiness of expression, the calculations are handwritten and attached as a file. Check the attached files for the complete calculation.

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Answer:

r = 4.24x10⁴ km.  

     

Explanation:

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From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

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I hope it helps you!

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How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
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Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

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Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.  

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Answer:

a lower energy than

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Explanation:

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