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Two neighbors return from a tropical vacation to find their houses at a frigid \(2^{\circ} \mathrm{C}\). Each house has a furnace that outputs \(10^{5} \mathrm{Btu} / \mathrm{h}\). One house is made of steel and has mass \(75,000 \mathrm{~kg}\) the other of wood with mass \(15,000 \mathrm{~kg}\). Neglecting heat loss, find the time required to bring each house to \(18^{\circ} \mathrm{C}\).

Short Answer

Expert verified
It takes approximately 5.23 hours to heat each house to 18°C.

Step by step solution

01

Understand the Heat Required

To determine the time required to heat each house, we first have to calculate the amount of heat energy needed to raise their temperatures from 2°C to 18°C. The formula to find heat required is \( Q = mc\Delta T \), where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. We'll calculate \( \Delta T \) first, which is \( 18^{\circ}C - 2^{\circ}C = 16^{\circ}C \).
02

Calculate Heat for Steel House

For the steel house, use the specific heat capacity of steel, \( c = 0.46 \text{ kJ/kg°C} \). First, convert \( 10^5 \text{ Btu/hr} \) to kJ, noting that \( 1 \text{ Btu} = 1.055 \text{ kJ} \). We find \( Q_\text{steel} = 75000 \times 0.46 \times 16 \). This results in \( Q_\text{steel} = 552000 \text{ kJ} \).
03

Calculate Time for Steel House

Find the energy output in kJ: \( 10^5 \text{ Btu/hr} = 105500 \text{ kJ/hr} \). The time \( t \) required is \( \frac{Q_\text{steel}}{\text{output}} = \frac{552000}{105500} \approx 5.23 \text{ hours} \).
04

Calculate Heat for Wood House

For the wood house, use the specific heat capacity of wood, \( c = 2.3 \text{ kJ/kg°C} \). Calculate heat: \( Q_\text{wood} = 15000 \times 2.3 \times 16 \). This results in \( Q_\text{wood} = 552000 \text{ kJ} \), coincidentally the same as for steel.
05

Calculate Time for Wood House

Again, use the energy output in kJ: \( 10^5 \text{ Btu/hr} = 105500 \text{ kJ/hr} \). The time \( t \) required is \( \frac{Q_\text{wood}}{\text{output}} = \frac{552000}{105500} \approx 5.23 \text{ hours} \).

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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, often simply called specific heat, is a property of materials that tells us how much heat energy is required to change the temperature of a given mass. It is defined as the heat required to raise the temperature of one kilogram of a material by one degree Celsius. The formula we use is:
  • \[Q = mc\Delta T\]
In this formula, \( Q \) represents the heat energy (in joules or kilojoules), \( m \) stands for mass (in kilograms), \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

Different materials have different specific heat capacities. For example, steel has a much lower specific heat than wood. This means that, for the same mass and temperature change, steel requires less energy to heat up compared to wood. In our problem, the specific heat capacity for steel is \( 0.46 \text{ kJ/kg°C} \) and for wood, it's \( 2.3 \text{ kJ/kg°C} \). These values significantly affect the calculations for how much heat is needed to increase the temperature of the houses.
Heat Transfer
Heat transfer is the process of thermal energy moving from a warmer object to a cooler one. In the context of heating homes, it's crucial to understand how heat is conducted through materials. We often use furnaces to provide a consistent source of heat energy, which transfers thermal energy evenly throughout a space.
  • Heat naturally moves from warmer to cooler areas.
  • The rate of heat transfer is influenced by material properties, the temperature difference, and the surface area involved.
In our problem, each house receives heat from a furnace outputting \( 10^5 \text{ Btu/hr} \), equivalent to \( 105500 \text{ kJ/hr} \). Understanding the relationship between heat input and the specific characteristics of materials is essential for efficient heating. The efficient transfer of this energy determines how quickly each house reaches the desired temperature of \( 18^{\circ}\text{C} \).

By calculating the energy required to heat each house, using their specific heat capacities, we understand that both houses require the same amount of heat energy due to the particular characteristics of their materials. However, the method and speed of heat transfer could differ if factored against insulation or the way different materials like wood and steel absorb heat.
Temperature Change
Temperature change, denoted as \( \Delta T \), refers to the difference in temperature between the initial state and the final state. In our exercise, the temperature change for each house is from \( 2^{\circ}C \) to \( 18^{\circ}C \).

Calculating \( \Delta T \) is a straightforward process:
  • The initial temperature (let's say, \( T_1 \)) is \( 2^{\circ}C \).
  • The final temperature (\( T_2 \)) desired is \( 18^{\circ}C \).
  • The temperature change \( \Delta T \) is simply \( T_2 - T_1 \).
  • In our case, \( \Delta T = 18^{\circ}C - 2^{\circ}C = 16^{\circ}C \).
This change helps us calculate the total heat energy needed to warm a particular material. Since both houses have the same \(\Delta T\), we can focus on differences in their mass and specific heat to understand their heating requirements.

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Most popular questions from this chapter

You're the public affairs director for an electric utility. A nuclear plant has been refueled and needs to power up again. The news media want to know how much time will pass before the reactor is on line again. The reactor needs to heat \(5.4 \times 10^{6} \mathrm{~kg}\) of water from \(10^{\circ} \mathrm{C}\) to \(350^{\circ} \mathrm{C}\). (It's a pressurized-water reactor, so the water remains liquid.) How much time will this take, if the reactor's thermal power output is \(1.42 \mathrm{GW} ?\) Ignore the heat capacity of the reactor vessel and plumbing.

A \(0.450-\mathrm{kg}\) rock is dropped \(10.0 \mathrm{~m}\) into \(2.5 \mathrm{~kg}\) of water. If all the rock's kinetic energy is converted to thermal energy in the water, what's the water's temperature increase?

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The top of a steel wood stove measures \(90 \mathrm{~cm}\) by \(40 \mathrm{~cm}\) If and is \(0.45 \mathrm{~cm}\) thick. The fire maintains the inside of the stovetop at \(310^{\circ} \mathrm{C}\), while the outside surface is \(295^{\circ} \mathrm{C}\). (a) Find the rate of heat conduction through the stovetop. (b) Suppose that the entire stove heats the air in the room at three times the rate you found in part (a). The room measures \(8.6 \mathrm{~m}\) by \(6.5 \mathrm{~m}\) by \(2.8 \mathrm{~m}\). Assuming the room is at \(T=20^{\circ} \mathrm{C}\) and \(P=1 \mathrm{~atm},\) at what rate does the room's temperature increase?

If Find the daily energy intake in joules for (a) an individual on a 1500-Cal/day weight-loss diet; (b) an athletic individual who consumes \(2600 \mathrm{Cal} /\) day; \((\mathrm{c})\) a Tour de France competitor consuming 6000 Cal/day.

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