Problem 64 - Physics

Question

To help keep his barn warm on cold days, a farmer stores \(840 \mathrm{~kg}\) of solar-heated water \(\left(L_{\mathrm{f}}=3.35 \times 10^{5} \mathrm{~J} / \mathrm{kg}\right)\) in barrels. For how many hours would a \(2.0-\mathrm{kW}\) electric space heater have to operate to provide the same amount of heat as the water does when it cools from 10.0 to \(0.0^{\circ} \mathrm{C}\) and completely freezes?

Step-by-Step Solution

Answer

The heater must operate for about 44 hours.

Step 1: Calculate Heat Lost During Cooling

First, we calculate the heat lost when water cools from 10°C to 0°C using the formula \( Q = mc\Delta T \), where \( m = 840 \text{ kg} \), \( c = 4.18 \times 10^3 \text{ J/kg°C} \) (specific heat capacity of water), and \( \Delta T = 0°C - 10°C = -10°C \).\[ Q = 840 \times 4.18 \times 10^3 \times (-10) \]Calculating gives \( Q = -35,112,000 \text{ J} \) or \( Q = 35,112,000 \text{ J} \) since heat loss is taken as positive.

Step 2: Calculate Heat Lost During Freezing

Next, calculate the heat loss during the phase change from water to ice using the latent heat of fusion formula \( Q = mL_f \), where \( L_f = 3.35 \times 10^5 \text{ J/kg} \).\[ Q = 840 \times 3.35 \times 10^5 \]Calculating gives \( Q = 281,400,000 \text{ J} \).

Step 3: Total Heat Lost by Water

Sum the heat lost during cooling and freezing to find the total heat released by the water.\[ Q_{ ext{total}} = 35,112,000 + 281,400,000 \]This gives \( Q_{ ext{total}} = 316,512,000 \text{ J} \).

Step 4: Calculate Equivalent Time for Electric Heater

Use the power equation \( P = \frac{W}{t} \) to find the time required for the space heater, where \( P = 2.0 \text{ kW} = 2000 \text{ W} \) and \( W = 316,512,000 \text{ J} \).Rearrange the equation to solve for time:\[ t = \frac{W}{P} = \frac{316,512,000}{2000} \]Calculating gives \( t = 158,256 \text{ s} \).

Step 5: Convert Time from Seconds to Hours

Finally, convert the time from seconds to hours. Since there are 3600 seconds in an hour, we use the conversion:\[ t_{\text{hours}} = \frac{158,256}{3600} \]Calculating gives \( t_{\text{hours}} = 43.96 \text{ hours} \), approximately 44 hours.

Key Concepts

Latent Heat of Fusion

When a substance changes from one phase to another, like from water to ice, a specific amount of heat energy is involved. This energy is called the latent heat of fusion. It is defined as the amount of heat required to change 1 kg of a substance from solid to liquid or vice versa, without changing its temperature.
For our problem, the latent heat of fusion for water is given by the formula:

• \[ Q = m L_f \]

where

• \(Q\) is the heat energy required,
• \(m\) is the mass of the substance (water here),
• \(L_f\) is the latent heat of fusion.

For water, \(L_f\) is \(3.35 \times 10^5 \text{ J/kg}\).
This implies that this much energy per kg is either absorbed or released when moving between solid and liquid states, without a change in temperature. In our exercise, the water released a substantial amount of energy while freezing, which was captured using the \(L_f\) value given.

Specific Heat Capacity

The specific heat capacity is a measure of the amount of heat energy required to raise the temperature of 1 kg of a substance by 1°C. In the given problem, we're dealing with water's specific heat capacity, an important factor in determining how much heat the water loses as it cools.
The formula to calculate the heat lost or gained is:

• \[ Q = mc\Delta T \]

where

• \(Q\) is the heat energy,
• \(m\) is the mass of the substance,
• \(c\) is the specific heat capacity,
• \(\Delta T\) is the change in temperature.

For water, the specific heat capacity is \(4.18 \times 10^3\) J/kg°C.
This means it takes 4180 J to change the temperature of 1 kg of water by 1°C. Understanding this helps us see how energy balances occur in everyday physical processes like heating or cooling substances.

Phase Change

Phase change involves the transition of a substance from one physical state to another, such as solid-to-liquid or liquid-to-gas. During a phase change, the temperature of the substance remains constant as all the added or removed energy goes into changing the phase.
In our problem, the water undergoes a phase change as it freezes to form ice. Initially, the water cools from 10°C to 0°C, releasing energy at a rate determined by its specific heat capacity.
Once it reaches 0°C, it begins to freeze, meaning it transitions from the liquid phase to the solid phase.

• During this phase change, significant energy is released due to the latent heat of fusion.

This is crucial in our example, as the energy released during this change was used to calculate how long an electric heater would have to run to provide the same heating effect.

Electric Power

Electric power is the rate at which electrical energy is consumed or produced by an electric circuit. It is typically measured in watts (W) or kilowatts (kW), where 1 kW equals 1000 W. In this exercise, we want to determine how long a 2.0 kW electric space heater needs to run to provide the equivalent amount of heat produced by the freezing water.
The formula that relates power, energy, and time is:

• \[ P = \frac{W}{t} \]

Where:

• \(P\) is the power in watts,
• \(W\) is the energy in joules,
• \(t\) is the time in seconds.

By rearranging this formula, we can solve for time:

• \[ t = \frac{W}{P} \]

Thus, using the calculated total energy from the water's cooling and freezing, \(316,512,000 \text{ J}\), and knowing the power of the heater is 2000 W, the required runtime in seconds can be found. This highlights how energy calculations in physics problems relate directly to real-world applications.