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Chapter 18: Problem 33

A metal brick found in an excavation was sent to a testing lab for nondestructive identification. The lab weighed the sample brick and found its mass to be \(3.0 \mathrm{~kg} .\) The brick was heated to a temperature of \(3.0 \cdot 10^{2}{ }^{\circ} \mathrm{C}\) and dropped into an insulated copper calorimeter of mass 1.5 kg containing \(2.0 \mathrm{~kg}\) of water at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} .\) The final temperature at equilibrium was noted to be \(31.7^{\circ} \mathrm{C}\). By calculating the specific heat of the sample from this data, can you identify the brick's material?

Short Answer

Expert verified
Answer: To calculate the specific heat capacity of the metal brick, use the following equation: \(c_b=\frac{m_w c_w (T_f-T_{wi})}{m_b (T_{f}-T_{bi})}\), where \(m_w\) and \(m_b\) denote the masses of water and brick, \(c_w\) is the specific heat capacity of water, and \(T_{wi}\), \(T_{bi}\), and \(T_f\) are the initial temperatures of water and brick and the final equilibrium temperature, respectively. Once the specific heat capacity of the brick, \(c_b\), is found, it can be compared to the specific heat capacities of common metals to possibly identify the material.

Step by step solution

01

Calculate the heat gained by the water

Since we know the mass of water \(m_w = 2.0 kg\), the initial temperature \(T_{wi}=20^\circ\,C\), final temperature \(T_{f}=31.7^\circ\,C\), and the specific heat capacity of water \(c_w=4186\, J/(kg\cdot K)\), we can calculate the heat gained by water using the formula: \(Q_w=m_w c_w (T_f-T_{wi})\)
02

Calculate the heat lost by the brick

We have the mass of the metal brick \(m_b = 3.0\,kg\), initial temperature \(T_{bi}=300^\circ\,C\), and the final temperature \(T_{f}=31.7^\circ\,C\). Let the specific heat of the brick be \(c_b\). Therefore, the heat lost by the brick can be calculated as: \(Q_b=m_b c_b (T_{f}-T_{bi})\)
03

Equate the heat gained by water and heat lost by the brick

For the system to reach equilibrium, the heat gained by water will be equal to the heat lost by brick i.e., \(Q_w=Q_b\). Plugging in the values from steps 1 and 2, we get: \(m_w c_w (T_f-T_{wi}) = m_b c_b (T_{f}-T_{bi})\)
04

Calculate the specific heat of the brick

Solve for \(c_b\) in the equation above: \(c_b=\frac{m_w c_w (T_f-T_{wi})}{m_b (T_{f}-T_{bi})}\) Now, plug in the known values and solve for \(c_b\).
05

Identify the material of the brick

Once we find the specific heat capacity \(c_b\) of the brick, we can compare it to the specific heat capacities of common metals and try to identify the material. Note, however, that without a proper list of specific heat capacities of different materials or metals, we won't be able to identify the material directly. The found value may be used as an indication of which material it might be.

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Most popular questions from this chapter

Suppose you mix 7.00 L of water at \(2.00 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) with \(3.00 \mathrm{~L}\) of water at \(32.0^{\circ} \mathrm{C}\); the water is insulated so that no energy can flow into it or out of it. (You can achieve this, approximately, by mixing the two fluids in a foam cooler of the kind used to keep drinks cool for picnics.) The \(10.0 \mathrm{~L}\) of water will come to some final temperature. What is this final temperature?

Several days after the end of a snowstorm, the roof of one house is still completely covered with snow, and another house's roof has no snow cover. Which house is most likely better insulated?

You were lost while hiking outside wearing only a bathing suit. a) Calculate the power radiated from your body, assuming that your body's surface area is about \(2.00 \mathrm{~m}^{2}\) and your skin temperature is about \(33.0^{\circ} \mathrm{C} .\) Also, assume that your body has an emissivity of 1.00 . b) Calculate the net radiated power from your body when you were inside a shelter at \(20.0^{\circ} \mathrm{C}\). c) Calculate the net radiated power from your body when your skin temperature dropped to \(27.0^{\circ} \mathrm{C}\).

A cryogenic storage container holds liquid helium, which boils at \(4.2 \mathrm{~K}\). Suppose a student painted the outer shell of the container black, turning it into a pseudoblackbody, and that the shell has an effective area of \(0.50 \mathrm{~m}^{2}\) and is at \(3.0 \cdot 10^{2} \mathrm{~K}\). a) Determine the rate of heat loss due to radiation. b) What is the rate at which the volume of the liquid helium in the container decreases as a result of boiling off? The latent heat of vaporization of liquid helium is \(20.9 \mathrm{~kJ} / \mathrm{kg} .\) The density of liquid helium is \(0.125 \mathrm{~kg} / \mathrm{L}\).

A thermos bottle fitted with a piston is filled with a gas. Since the thermos bottle is well insulated, no heat can enter or leave it. The piston is pushed in, compressing the gas. a) What happens to the pressure of the gas? Does it increase, decrease, or stay the same? b) What happens to the temperature of the gas? Does it increase, decrease, or stay the same? c) Do any other properties of the gas change?
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