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Consider a hot baked potato. Will the potato cool faster or slower when we blow the warm air coming from our lungs on it instead of letting it cool naturally in the cooler air in the room? Explain.

Short Answer

Expert verified
Answer: Blowing warm air from our lungs onto a hot baked potato slows down its cooling process compared to letting it cool naturally in the cooler air in the room. This is because the temperature difference between the potato and the warm air is less than that between the potato and the cooler room air, resulting in a slower rate of heat transfer by convection.

Step by step solution

01

In this case, the heat transfer occurs when the temperature of the hot baked potato is higher than the air temperature around it. The heat energy from the potato will transfer to the cooler air in the room, making the potato cool down gradually. This is a process called convection. #Step 2: Analyze the effect of blowing warm air on the potato#

When we blow warm air from our lungs onto the potato, we are essentially placing the potato in a temporary environment where the air temperature is higher than that of the room air. Therefore, the temperature difference between the potato and the air blown from our lungs will be less than that between the potato and the room air. #Step 3: Compare heat transfer rates#
02

Heat transfer by convection is directly proportional to the temperature difference between the hot object (in this case, the potato) and the surrounding air. When the surrounding air temperature is closer to that of the hot object (as it is when we blow warm air on it), the rate of heat transfer will be slower compared to when the surrounding air temperature is cooler in the room. #Step 4: Explain the observed effect on cooling rate#

Since the rate of heat transfer is slower when we blow warm air on the potato, it can be concluded that the potato will cool slower when we blow warm air on it as compared to letting it cool naturally in the cooler air in the room.

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Most popular questions from this chapter

The upper surface of a \(1-\mathrm{m} \times 1-\mathrm{m}\) ASTM B 152 copper plate is being cooled by air at \(20^{\circ} \mathrm{C}\). The air is flowing parallel over the plate surface at a velocity of $0.5 \mathrm{~m} / \mathrm{s}$. The local convection heat transfer coefficient along the plate surface is correlated with the correlation \(h_{x}=1.36 x^{-0.5}\), where \(h_{x}\) and \(x\) have the units \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{m}\), respectively. The maximum use temperature for the ASTM B152 copper plate is \(260^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-1M). Evaluate the average convection heat transfer coefficient \(h\) for the entire plate. If the rate of convection heat transfer from the plate surface is \(700 \mathrm{~W}\), would the use of ASTM B152 plate be in compliance with the ASME Code for Process Piping?

Two metal plates are connected by a long ASTM A479 904L stainless steel bar. A hot gas, at \(400^{\circ} \mathrm{C}\), flows between the plates and across the bar. The bar has a square cross section with a width of \(2 \mathrm{~cm}\), and the length of the bar exposed to the hot gas is \(10 \mathrm{~cm}\). The average convection heat transfer coefficient for the bar in crossflow is correlated with the gas velocity as \(h=13.6 V^{0.675}\), where \(h\) and \(V\) have the units \(\mathrm{W} / \mathrm{m}^{2}, \mathrm{~K}\) and \(\mathrm{m} / \mathrm{s}\), respectively. The maximum use temperature for the ASTM A479 904L is \(260^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-1M). The temperature of the bar is maintained by a cooling mechanism with the capability of removing heat at a rate of 100 W. Determine the maximum velocity that the gas can achieve without heating the stainless steel bar above the maximum use temperature set by the ASME Code for Process Piping.

What does the friction coefficient represent in flow over a flat plate? How is it related to the drag force acting on the plate?

Air at \(5^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), is used for cooling metal plates coming out of a heat treatment oven at an initial temperature of \(300^{\circ} \mathrm{C}\). The plates $\left(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\right.$, and \(\left.c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) have a thickness of \(10 \mathrm{~mm}\). Using appropriate software, determine the effect of cooling time on the temperature gradient in the metal plates at the surface. By varying the cooling time from 0 to \(3000 \mathrm{~s}\), plot the temperature gradient in the plates at the surface as a function of cooling time. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

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