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Chapter 1: Problem 149

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 minutes. 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: (c) 58 W

Step by step solution

01

Calculate the change in temperature

First, we need to find the change in temperature (\(\Delta T\)) of the drink. The difference between the initial temperature (\(T_i = 5^{\circ}C\)) and the final temperature (\(T_f = 15^{\circ}C\)) is \(\Delta T = T_f - T_i\). \(\Delta T = 15^{\circ}C - 5^{\circ}C = 10^{\circ}C\)
02

Compute the heat gained by the drink

Now, we need to find the heat gained by the drink. We will use the heat formula: \(Q = m \cdot c_p \cdot \Delta T\). \(Q = (2.5 ~kg) \cdot (4200 ~ J / kg \cdot K) \cdot (10 ~K)\) \(Q = 105000 ~J\)
03

Convert the time taken to seconds

We are given that the time taken for the temperature to rise is 30 minutes. We need to convert this time to seconds: \(t = 30 ~minutes \cdot 60 ~s / minute = 1800 ~s\).
04

Calculate the average rate of heat transfer

Now, we can find the average rate of heat transfer, which is the power. We'll use the formula for power: \(P = \frac{Q}{t}\). \(P = \frac{105000 ~ J}{1800 ~s} = 58.33 ~ W\)
05

Choose the correct answer

The average rate of heat transfer to the drink is closest to answer (c) \(58 ~ W\).

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Most popular questions from this chapter

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 equivalent air temperature equal 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 \(\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 V^{0.16}+0.3965 T V^{0.16}\) 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.

A room is heated by a \(1.2 \mathrm{~kW}\) electric resistance heater whose wires have a diameter of \(4 \mathrm{~mm}\) and a total length of \(3.4 \mathrm{~m}\). The air in the room is at \(23^{\circ} \mathrm{C}\) and the interior surfaces of the room are at \(17^{\circ} \mathrm{C}\). The convection heat transfer coefficient on the surface of the wires is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the rates of heat transfer from the wires to the room by convection and by radiation are equal, the surface temperature of the wire is (a) \(3534^{\circ} \mathrm{C}\) (b) \(1778^{\circ} \mathrm{C}\) (c) \(1772^{\circ} \mathrm{C}\) (d) \(98^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

Determine a positive real root of this equation using \(E E S\) : $$ 3.5 x^{3}-10 x^{0.5}-3 x=-4 $$

On a still clear night, the sky appears to be a blackbody with an equivalent temperature of \(250 \mathrm{~K}\). What is the air temperature when a strawberry field cools to \(0^{\circ} \mathrm{C}\) and freezes if the heat transfer coefficient between the plants and air is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) because of a light breeze and the plants have an emissivity of \(0.9\) ? (a) \(14^{\circ} \mathrm{C}\) (b) \(7^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-3^{\circ} \mathrm{C}\)

A 40-cm-long, 0.4-cm-diameter electric resistance wire submerged in water is used to determine the convection heat transfer coefficient in water during boiling at \(1 \mathrm{~atm}\) pressure. The surface temperature of the wire is measured to be \(114^{\circ} \mathrm{C}\) when a wattmeter indicates the electric power consumption to be \(7.6 \mathrm{~kW}\). The heat transfer coefficient is (a) \(108 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(13.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(68.1 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(0.76 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(256 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)
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