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Chapter 12: Problem 43

Review Interactive Solution \(\underline{12.43}\) at for help in approaching this problem. When resting, a person has a metabolic rate of about \(3.0 \times 10^{5}\) joules per hour. The person is submerged neck-deep into a tub containing \(1.2 \times 10^{3} \mathrm{~kg}\) of water at \(21.00{ }^{\circ} \mathrm{C}\). If the heat from the person goes only into the water, find the water temperature after half an hour.

Short Answer

Expert verified
The final water temperature is approximately \(21.03^\circ \text{C}\).

Step by step solution

01

Understand the Problem

We are tasked with finding the temperature change of water after a person, with a given metabolic rate, transfers heat to it over a specified period. We need to consider the heat transfer from the person to the water and how that affects the water's temperature.
02

Calculate Total Heat Transferred

The metabolic rate is given as \(3.0 \times 10^5\) joules per hour. Since the person is submerged for half an hour, the total heat transferred \(Q\) to the water is calculated by multiplying the rate by the time: \[ Q = 3.0 \times 10^5 \, \text{J/h} \times 0.5 \, \text{h} = 1.5 \times 10^5 \, \text{J}. \]
03

Use the Heat Transfer Formula

The formula for heat transfer, based on mass \(m\), specific heat capacity \(c\), and temperature change \(\Delta T\), is \(Q = mc\Delta T\). For water, \(m = 1.2 \times 10^3 \, \text{kg}\) and \(c = 4.186 \text{ J/(g}^\circ \text{C)} = 4.186 \times 10^3 \text{ J/(kg}^\circ \text{C)}\).
04

Solve for Temperature Change

Rearrange the heat transfer formula to solve for \(\Delta T\): \[ \Delta T = \frac{Q}{mc} = \frac{1.5 \times 10^5}{1.2 \times 10^3 \times 4.186 \times 10^3}. \] Calculate \(\Delta T\): \[ \Delta T \approx \frac{1.5 \times 10^5}{5.0232 \times 10^6} \approx 0.0299^\circ \text{C}. \]
05

Calculate Final Water Temperature

Add the temperature change to the initial water temperature to find the final temperature: \[ 21.00^\circ \text{C} + 0.0299^\circ \text{C} \approx 21.03^\circ \text{C}. \]

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Key Concepts

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

Metabolic Rate
The metabolic rate is the rate at which a person's body converts energy stored in food into necessary functions, such as maintaining body temperature. When we say a person has a metabolic rate of \(3.0 \times 10^5\) joules per hour, we mean that their body is releasing energy at this rate.

This energy can be converted in many ways, such as movement or heat. In this exercise, the energy is transferred as heat to water, causing a temperature change. Metabolic rate influences how much energy is available to be transferred over time. This is why we need to consider how long the person is submerged in the water to estimate total heat transfer.

In our example, over half an hour, 150,000 joules of energy is transferred to the water.
Specific Heat Capacity
Specific heat capacity is a property of a material that indicates how much heat energy is required to raise the temperature of a unit mass of the material by one degree Celsius. It is denoted as \(c\).

The specific heat capacity of water is \(4.186 \times 10^3 \text{ J/(kg}^\circ \text{C)}\). This relatively high value means water requires a lot of energy to change its temperature.

In this problem, we use the specific heat capacity value to understand how much heat energy from the metabolic transfer affects the water鈥檚 temperature. Knowing the specific heat helps predict how substances like water respond to heat input.
Temperature Change
Temperature change \(\Delta T\) is the variation in the temperature of a substance as it absorbs or loses energy. It is calculated by rearranging the heat transfer formula:
  • \(Q = mc\Delta T\)
  • Rewrite as: \(\Delta T = \frac{Q}{mc}\)

For this problem, the heat \(Q\) is the energy transferred from the person to the water. With the mass \(m\) of \(1.2 \times 10^3 \text{ kg}\) and the specific heat \(c\), we calculate how much the temperature of the water has increased due to this energy exchange.

The small value \(\Delta T \approx 0.0299^\circ C\) shows that even with considerable energy transfer, the temperature rise in water is minimal, given water's high specific heat capacity.
Joules
Joules are the unit of energy in the International System of Units (SI), and are used to quantify energy transfer or work done. One joule is equivalent to the energy transferred when a force of one newton acts over a distance of one meter.

In this context, joules represent the heat energy transferred from the person to the water. The metabolic rate of \(3.0 \times 10^5\) joules per hour indicates the rate of energy release, while the computation of \(1.5 \times 10^5\) joules is the total energy transferred over half an hour.

This setup allows us to calculate how many joules are applied to the water to determine its temperature change.

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Most popular questions from this chapter

A rock of mass \(0.20 \mathrm{~kg}\) falls from rest from a height of \(15 \mathrm{~m}\) into a pail containing \(0.35 \mathrm{~kg}\) of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is \(1840 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). Ignore the heat absorbed by the pail itself, and determine the rise in the temperature of the rock and water.

Multiple-Concept Example 4 reviews the concepts that are involved in this problem. A ruler is accurate when the temperature is \(25^{\circ} \mathrm{C}\). When the temperature drops to \(-14^{\circ} \mathrm{C}\), the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} \mathrm{~N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross-sectional area of \(1.6 \times 10^{-5} \mathrm{~m}^{2},\) and it is made from a material whose coefficient of linear expansion is \(2.5 \times 10^{-5}\left(\mathrm{C}^{0}\right)^{-1}\). What is Young's modulus for the material from which the ruler is made?

To help prevent frost damage, fruit growers sometimes protect their crop by spraying it with water when overnight temperatures are expected to go below the freezing mark. When the water turns to ice during the night, heat is released into the plants, thereby giving them a measure of protection against the falling temperature. Suppose a grower sprays \(7.2 \mathrm{~kg}\) of water at \(0^{\circ} \mathrm{C}\) onto a fruit tree. (a) How much heat is released by the water when it freezes? (b) How much would the temperature of a \(180-\mathrm{kg}\) tree rise if it absorbed the heat released in part (a)? Assume that the specific heat capacity of the tree is \(2.5 \times 10^{3} \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and that no phase change occurs within the tree itself.

On a cool autumn morning, the relative humidity is \(100 \%\). (a) Does this mean that the partial pressure of water in the air equals atmospheric pressure? Why or why not? (b) Suppose the air warms up in the afternoon, but the partial pressure of water in the air does not change. Is the humidity still \(100 \%\) ? Provide a reason for your answer. The vapor pressure of water at \(20{ }^{\circ} \mathrm{C}\) is \(2500 \mathrm{~Pa}\). (a) What percentage of atmospheric pressure is this? (b) What percentage of the total air pressure at \(20{ }^{\circ} \mathrm{C}\) is due to water vapor if the relative humidity is \(100 \%\) ? (c) The vapor pressure of water at \(35^{\circ} \mathrm{C}\) is \(5500 \mathrm{~Pa}\). What is the relative humidity at this temperature if the partial pressure of water in the air has not changed from that at \(20{ }^{\circ} \mathrm{C}\) ?

A person eats a container of strawberry yogurt. The Nutritional Facts label states that it contains 240 Calories ( 1 Calorie \(=4186 \mathrm{~J}\) ). What mass of perspiration would one have to lose to get rid of this energy? At body temperature, the latent heat of vaporization of water is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\)
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