Problem 43 - Physics

Question

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.

Step-by-Step Solution

Answer

The water temperature after half an hour is approximately 21.03°C.

Step 1: Understanding the Problem

The problem involves calculating the change in temperature of water caused by the heat energy transferred from a person to the water. We know the metabolic rate of the person and the mass of the water.

Step 2: Calculate Heat Energy Produced by the Person

The metabolic rate of the person is given as \(3.0 \times 10^{5}\) joules per hour. In half an hour, the person produces \(\frac{3.0 \times 10^{5}}{2} = 1.5 \times 10^{5}\) joules of energy.

Step 3: Use the Heat Energy to Calculate Temperature Change

The specific heat capacity of water is \(c = 4186 \, \text{J/kg}^\circ\text{C}\). The formula to calculate the temperature change is \(Q = mc\Delta T\), where \(Q\) is the heat absorbed, \(m\) is the mass of the water, and \(\Delta T\) is the change in temperature. Rearranging gives \(\Delta T = \frac{Q}{mc}\).

Step 4: Substitute Values into the Formula

Substitute the values into the formula: \(\Delta T = \frac{1.5 \times 10^{5}}{1.2 \times 10^{3} \times 4186}\).

Step 5: Compute the Change in Temperature

Calculate \(\Delta T: \Delta T = \frac{1.5 \times 10^{5}}{5023200} \approx 0.0298\).

Step 6: Determine Final Water Temperature

Add the calculated temperature change to the initial water temperature: \(21.00 + 0.0298 \approx 21.03^{\circ}\text{C}\).

Key Concepts

Metabolic Rate

When looking at heat transfer, the metabolic rate is crucial in understanding how much energy a human body produces over time. The metabolic rate refers to the rate at which a body converts chemical energy from food into mechanical energy or heat. In the context of our problem, the person's metabolic rate is given as \(3.0 \times 10^{5}\) joules per hour. This means the body continuously generates 300,000 joules of energy every hour, which is then either released as heat or used up by bodily functions.

To calculate the energy produced in a shorter time, such as half an hour, we simply adjust the rate accordingly. For half an hour, we calculate:

• Energy produced = \(\frac{3.0 \times 10^{5}}{2} = 1.5 \times 10^{5}\) joules.

As a result, the body transfers this exact amount of energy as heat into the surrounding water during the mentioned time frame.

Specific Heat Capacity

The specific heat capacity is an essential concept in heat transfer studies. It tells us how much heat a substance requires to change its temperature by one degree Celsius. For water, this value is quite high, specifically, \(4186 \text{ J/kg}^\circ\text{C}\). This means water needs 4186 joules of heat to raise the temperature of just one kilogram by one degree Celsius. This is why water is so effective at absorbing heat, making it ideal for maintaining or changing temperatures in various situations.

In our problem, knowing the specific heat capacity allows us to predict how much the temperature of the water will change when a certain amount of energy is added. Because water has a high specific heat, it absorbs a lot of heat without a significant temperature change, which is why the temperature rise is quite minimal despite the input of substantial energy.

Temperature Change

Temperature change in the context of heat transfer is the measure of how much a substance's temperature increases or decreases after absorbing or releasing heat. In this exercise, we calculate the temperature change by using the formula:

• \(Q = mc\Delta T\)

Here, \(Q\) represents the heat energy transferred, \(m\) stands for the mass of the water, and \(c\) is the specific heat capacity. Our goal is to find \(\Delta T\), the change in temperature.

By rearranging the formula to solve for \(\Delta T\), we get:

• \(\Delta T = \frac{Q}{mc}\)

Substituting in the given values:

• \(\Delta T = \frac{1.5 \times 10^{5}}{1.2 \times 10^{3} \times 4186} \approx 0.0298 \text{ degrees Celsius}\).

This minute change demonstrates the efficiency with which water retains its temperature when subjected to heat input.

Energy Transfer

Energy transfer is a fundamental process seen in various physical phenomena and is central to our exercise. It refers to the movement of energy from one body or system to another. In the problem solved, energy is specifically transferred from a person to water. The body, through metabolic processes, generates heat energy. This energy is then moved from the body to the water, raising its temperature minimally.

The overall process can be broken down into straightforward steps:

• Calculate energy produced by the person, which is provided as \(1.5 \times 10^{5}\) joules in half an hour.
• Determine how this energy affects the water temperature by considering its mass and specific heat capacity.

Mastering energy transfer calculations help us predict and understand how various environments alter their surroundings, whether cooling down or heating up, based on the energy they emit or absorb.