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A cold bottled drink ( $\left.m=2.5 \mathrm{~kg}, c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)5^{\circ} \mathrm{C}$ is left on a table in a room. The average temperature of the drink is observed to rise to \(15^{\circ} \mathrm{C}\) in \(30 \mathrm{~min}\). The average rate of heat transfer to the drink is (a) \(23 \mathrm{~W}\) (b) \(29 \mathrm{~W}\) (c) \(58 \mathrm{~W}\) (d) \(88 \mathrm{~W}\) (e) \(122 \mathrm{~W}\)

Short Answer

Expert verified
Answer: The average rate of heat transfer to the drink is 58 W.

Step by step solution

01

Calculate the Heat Gained

We will first calculate the Heat gained(Q) by the drink using the formula Q = mcΔT, where m is the mass, c is the specific heat, and ΔT = Tf - Ti is the change in temperature. ΔT = Tf - Ti = \((15 - 5)^\circ \mathrm{C} = 10^\circ \mathrm{C}\) Now, calculate the heat gained(Q): \(Q = m \cdot c_{p} \cdot ΔT = 2.5 \mathrm{~kg} \cdot 4200 \mathrm{~J}/(\mathrm{kg} \cdot \mathrm{K}) \cdot 10 \mathrm{~K} = 105000 \mathrm{~J}\)
02

Calculate the Average Rate of Heat Transfer

To find the average rate of heat transfer, we will use the formula for average power(P) and time(t), which is given in minutes. We need to convert the time to seconds for calculations. Convert the time to seconds: \(t = 30 \mathrm{~min} \cdot 60 \mathrm{~s/min} = 1800 \mathrm{~s}\) Now, calculate the average rate of heat transfer(P): \(P = \frac{Q}{t} = \frac{105000 \mathrm{~J}}{1800 \mathrm{~s}} = 58.3 \mathrm{~W}\) The answer is closest to option (c), so the average rate of heat transfer to the drink is 58 W.

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Most popular questions from this chapter

An electric current of 5 A passing through a resistor has a measured voltage of \(6 \mathrm{~V}\) across the resistor. The resistor is cylindrical with a diameter of \(2.5 \mathrm{~cm}\) and length of \(15 \mathrm{~cm}\). The resistor has a uniform temperature of \(90^{\circ} \mathrm{C}\), and the room air temperature is \(20^{\circ} \mathrm{C}\). Assuming that heat transfer by radiation is negligible, determine the heat transfer coefficient by convection.

A 1000-W iron is left on an ironing board with its base exposed to the air at \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the base surface and the surrounding air is $35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. If the base has an emissivity of \)0.6$ and a surface area of \(0.02 \mathrm{~m}^{2}\), determine the temperature of the base of the iron. Answer: \(674^{\circ} \mathrm{C}\)

It is well known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind-chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or $5 \mathrm{~km} / \mathrm{h}$ ). The hypothetical wind-chill temperature (WCT), called the wind-chill temperature index (WCTI), is an air temperature equivalent to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind-chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI $\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 \mathrm{~V}^{0.16}+0.3965 T V^{0.16}$ where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI $\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}$ where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite

An electric heater with the total surface area of \(0.25 \mathrm{~m}^{2}\) and emissivity \(0.75\) is in a room where the air has a temperature of $20^{\circ} \mathrm{C}\( and the walls are at \)10^{\circ} \mathrm{C}$. When the heater consumes \(500 \mathrm{~W}\) of electric power, its surface has a steady temperature of \(120^{\circ} \mathrm{C}\). Determine the temperature of the heater surface when it consumes \(700 \mathrm{~W}\). Solve the problem (a) assuming negligible radiation and (b) taking radiation into consideration. Based on your results, comment on the assumption made in part \((a)\).

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be $25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and disregard any heat loss by radiation.

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