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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Specific heat capacity is a crucial concept in understanding how different materials respond to heat. It tells us how much heat energy is needed to change the temperature of a given mass by one degree Celsius. The formula for specific heat capacity is:\[ c_p = \frac{Q}{m \cdot \Delta T} \]where:
  • \(Q\) is the amount of heat added (in Joules).
  • \(m\) is the mass of the substance (in kilograms).
  • \(\Delta T\) is the change in temperature (in Celsius or Kelvin).
In the context of our problem, the specific heat capacity of the drink is given as 4200 J/kg路K. This implies that each kilogram of the drink requires 4200 Joules to increase its temperature by 1 degree Celsius. Understanding this helps us determine how much energy was gained by the drink as it warmed from 5掳C to 15掳C.
Thermal Analysis
Thermal analysis involves examining how heat flows into and out of a system, leading to changes in temperature. This is vital in many real-life applications, from designing heating systems to understanding how climate control works.In our problem, we conduct a thermal analysis to find how much heat energy the drink absorbed as it warmed up. We utilize the formula:\[ Q = m \cdot c_p \cdot \Delta T \]Here, the drink absorbs \(105000 ~ J\) as it warms by \(10^{\circ}C\) with a mass of 2.5 kg. By calculating the energy transfer, we gain insight into the thermal behavior of the drink under ambient conditions. Moreover, such analyses help link physical conditions to energy implications, which is fundamental in engineering and environmental science.
Rate of Heat Transfer
The rate of heat transfer is a measure of how quickly heat energy moves from one place to another. It is often expressed in watts (W), where one watt is equal to one joule per second.For our drink, after calculating the total heat absorbed, which was \(105000 ~ J\), we determined the time it took for this heat to be absorbed, which is 1800 seconds. The rate of heat transfer is calculated using the formula:\[ P = \frac{Q}{t} \]where:
  • \(P\) is power in watts.
  • \(Q\) is the total heat energy in Joules.
  • \(t\) is the time in seconds.
For our case, the average rate of heat transfer is \(58.33 ~ W\), indicating the speed at which the surrounding air heated the drink. Recognizing this rate is essential for designing cooling and heating systems efficiently.

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Most popular questions from this chapter

Water enters a pipe at \(20^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{~kg} / \mathrm{s}\) and is heated to \(60^{\circ} \mathrm{C}\). The rate of heat transfer to the water is (a) \(20 \mathrm{~kW}\) (b) \(42 \mathrm{~kW}\) (c) \(84 \mathrm{~kW}\) (d) \(126 \mathrm{~kW}\) (e) \(334 \mathrm{~kW}\)

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

An ice skating rink is located in a building where the air is at \(T_{\text {air }}=20^{\circ} \mathrm{C}\) and the walls are at \(T_{w}=25^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\) and its density is \(920 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

The inner and outer surfaces of a \(25-\mathrm{cm}\)-thick wall in summer are at \(27^{\circ} \mathrm{C}\) and \(44^{\circ} \mathrm{C}\), respectively. The outer surface of the wall exchanges heat by radiation with surrounding surfaces at \(40^{\circ} \mathrm{C}\), and convection with ambient air also at \(40^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\). If both the emissivity and the solar absorptivity of the outer surface are \(0.8\), determine the effective thermal conductivity of the wall.

Solve this system of three equations with three unknowns using EES: $$ \begin{array}{r} x^{2} y-z=1.5 \\ x-3 y^{0.5}+x z=-2 \\ x+y-z=4.2 \end{array} $$
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