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Chapter 1: Problem 87

Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of \(500-^{\circ} \mathrm{C}\) granite must be placed in \(4.00 \mathrm{kg}\) of \(15.0-^{\circ} \mathrm{C}\) water to bring its temperature to \(100^{\circ} \mathrm{C}\), if \(0.0250 \mathrm{kg}\) of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings.

Short Answer

Expert verified
The mass of granite required to bring the water temperature to \(100^{\circ}C\) is approximately \(0.183 kg\).

Step by step solution

01

Identify the specific heat values

Specific heat capacity of water is given by \(c_{water} = 4186 \frac{J}{kg \cdot K}\), the specific heat capacity of granite is given by \(c_{granite} = 790 \frac{J}{kg \cdot K}\), and the heat of vaporization for water is given by \(L_v = 2.26 \times 10^6 \frac{J}{kg}\).
02

Write down the masses and temperatures of water and granite

Mass of water initially \(m_{water} = 4.00 kg\), and mass of the water which escapes as vapor is \(m_{escape} = 0.0250 kg\). Therefore, the mass of water remaining after initial sizzle is \(m_{waterRem} = m_{water} - m_{escape} = 3.975 kg\). The initial temperature of water is \(T_{waterInitial} = 15.0 ^{\circ}C = 288.15K\), the final temperature of water is \(T_{waterFinal} = 100^{\circ}C = 373.15K\), and the initial temperature of the granite is \(T_{graniteInitial} = 500^{\circ}C = 773.15K\). We need to find the mass of granite, \(m_{granite}\), which has to be placed in the water to bring its temperature to \(100^{\circ}C\).
03

Apply the conservation of energy principle

Conservation of energy states that energy gained by the water should equal the energy lost by the granite and the energy used to vaporize water. Energy gained by water: \[Q_{water} = m_{waterRem} c_{water} (T_{waterFinal} - T_{waterInitial})\] Energy used to vaporize water: \[Q_{vapor} = m_{escape} L_v\] Energy lost by granite: \[Q_{granite} = m_{granite} c_{granite} (T_{graniteInitial} - T_{waterFinal})\] The conservation of energy equation can be written as: \[Q_{water} + Q_{vapor} = Q_{granite}\]
04

Solve the conservation of energy equation for mass of granite

Substitute the values from steps 1, 2 and 3, into the equation and solve for mass of granite: \(m_{waterRem} c_{water} (T_{waterFinal} - T_{waterInitial}) + m_{escape} L_v = m_{granite} c_{granite} (T_{graniteInitial} - T_{waterFinal})\) Rearrange the equation for \(m_{granite}\): \(m_{granite} = \frac{m_{waterRem} c_{water} (T_{waterFinal} - T_{waterInitial}) + m_{escape} L_v}{c_{granite} (T_{graniteInitial} - T_{waterFinal})}\) Plug in the known values: \(m_{granite} = \frac{3.975kg \times 4186 \frac{J}{kg \cdot K} \times (373.15K - 288.15K) + 0.0250kg \times 2.26 \times 10^6 \frac{J}{kg}}{790 \frac{J}{kg \cdot K} \times (773.15K - 373.15K)}\) Calculate the mass of granite: \(m_{granite} \approx 0.183 kg\)
05

Final Answer

The mass of granite required to bring the water temperature to \(100^{\circ}C\) is approximately \(0.183 kg\).

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Key Concepts

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

Conservation of Energy
The conservation of energy is a fundamental principle in physics stating that energy cannot be created or destroyed, only transferred or transformed. In thermodynamics, this principle is essential for understanding how systems such as the one in our exercise balance energy.

When we use this principle with the granite and water system, the energy lost by the granite due to its cooling must be equal to the energy gained by the water, plus the energy used to vaporize some water. Granite, initially at a high temperature, transfers its energy to the water until thermal equilibrium is reached.
  • Energy gained by the water raises its temperature from 15掳C to 100掳C.
  • Energy lost by the granite decreases its temperature from 500掳C to 100掳C.
  • A part of this transferred energy also changes the state of some water from liquid to steam.
In summary, this energy balance ensures that all energy that leaves one part of the system enters another, keeping the total energy constant.
Specific Heat Capacity
Specific heat capacity is a property of material that indicates the amount of heat required to change the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). It is a measure of how much energy a material can store or absorb. This concept is critical in solving our exercise.

In our problem, different materials (water and granite) have unique specific heat capacities:
  • Water: 4186 J/kg路K 鈥� Water is excellent at storing heat, which is why it takes a lot of energy to change its temperature.
  • Granite: 790 J/kg路K 鈥� Granite requires less energy to change its temperature, making it efficient for different thermal applications.
Knowing these values helps calculate how much energy is required to change the water's temperature and how much energy the granite can provide by cooling. Using these specific heat values in calculations, we can predict the thermal interactions in this system and find the required mass of granite.
Heat of Vaporization
Heat of vaporization is the quantity of heat needed to convert a unit mass of liquid into vapor without a temperature change. For water, this value is quite high, reflecting the significant energy required for phase changes from liquid water to steam.

In this exercise, some of the initial water vaporizes, meaning that some energy from the granite is used not to just raise the temperature but to change the liquid water into vapor. The heat of vaporization for water is 2.26 x 10^6 J/kg, which is substantial. This means that even a small amount of water turning into vapor can take away a considerable amount of energy.
  • This energy requirement is vital for calculating the conservation of energy.
  • It ensures that the energy loss through vaporization is accounted for in the total energy calculation.
Understanding heat of vaporization is essential for solving thermodynamic problems where phase changes occur, as seen in our scenario where steam is produced under heating.

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Most popular questions from this chapter

Your house will be empty for a while in cold weather, and you want to save energy and money. Should you turn the thermostat down to the lowest level that will protect the house from damage such as freezing pipes, or leave it at the normal temperature? (If you don't like coming back to a cold house, imagine that a timer controls the heating system so the house will be warm when you get back.) Explain your answer.

How is heat transfer related to temperature?

If you pour 0.0100 kg of \(20.0^{\circ} \mathrm{C}\) water onto a \(\left.1.20-\mathrm{kg} \text { block of ice (which is initially at }-15.0^{\circ} \mathrm{C}\right),\) what is the final temperature? You may assume that the water cools so rapidly that effects of the surroundings are negligible.

In a study of healthy young men \(^{[11, \text { doing } 20\text { push- } }\) ups in 1 minute burned an amount of energy per \(\mathrm{kg}\) that for a 70.0 -kg man corresponds to 8.06 calories (kcal). How much would a 70.0 -kg man's temperature rise if he did not lose any heat during that time?

A pendulum is made of a rod of length \(L\) and negligible mass, but capable of thermal expansion, and a weight of negligible size. (a) Show that when the temperature increases by \(d T,\) the period of the pendulum increases by a fraction \(\alpha L d T / 2 .\) (b) A clock controlled by a brass pendulum keeps time correctly at \(10^{\circ} \mathrm{C}\). If the room temperature is \(30^{\circ} \mathrm{C}\), does the clock run faster or slower? What is its error in seconds per day?
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