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Chapter 18: Problem 40

Suppose \(0.010 \mathrm{~kg}\) of steam (at \(100.00^{\circ} \mathrm{C}\) ) is added to \(0.10 \mathrm{~kg}\) of water (initially at \(\left.19.0^{\circ} \mathrm{C}\right)\). The water is inside an aluminum cup of mass \(35 \mathrm{~g}\). The cup is inside a perfectly insulated calorimetry container that prevents heat flow with the outside environment. Find the final temperature of the water after equilibrium is reached.

Short Answer

Expert verified
Answer: The final equilibrium temperature of the water after steam is added is approximately \(63.84^{\circ} \mathrm{C}\).

Step by step solution

01

Calculate the Heat Gained by the Water and Aluminum Cup

First, we need the specific heat capacity of water \(c_w = 4.18 \times 10^3 \, \mathrm{J/kg \cdot K} \) and aluminum, \(c_{Al} = 0.897 \times 10^3 \, \mathrm{J/kg \cdot K} \). Let the initial temperature of water and the aluminum cup be \(T_1\), the equilibrium temperature be \(T_f\) and let the mass of water and aluminum cup are \(m_w\) and \(m_{Al}\), respectively. The heat absorbed by the water and the aluminum cup can be written as follows: $$ Q_{water} = m_w \cdot c_w \cdot (T_f - T_1)$$ $$ Q_{Al} = m_{Al} \cdot c_{Al} \cdot (T_f - T_1)$$ Now, let's move on to step 2.
02

Calculate the Heat Released by the Steam

We are given the mass of steam \(m_s = 0.010 \,\mathrm{kg}\). When steam condenses to become water at \(100^{\circ}\mathrm{C}\), it releases heat. The heat of vaporization of water is \(L_v = 2.26 \times 10^6 \, \mathrm{J/kg}\). So, the heat released during this process can be calculated as follows: $$ Q_{steam} = m_s \cdot L_v $$ In addition, we need to calculate the heat released when the condensed water cools down to the equilibrium temperature, \(T_f\): $$ Q_{s_cooling} = m_s \cdot c_w \cdot (100 - T_f)$$ Now, we will move on to step 3.
03

Equate the Heat Gained and Heat Released

According to the conservation of energy principle, the heat gained by the water and the aluminum cup must be equal to the heat released by steam: $$ Q_{water} + Q_{Al} = Q_{steam} + Q_{s_cooling}$$ Now, plug in the values calculated in steps 1 and 2: $$ m_w \cdot c_w \cdot (T_f - T_1) + m_{Al} \cdot c_{Al} \cdot (T_f - T_1) = m_s \cdot L_v + m_s \cdot c_w \cdot (100 - T_f) $$
04

Solve for the Final Equilibrium Temperature

Now, we can plug in the known values (\(m_w, m_s, m_{Al}, c_w, c_{Al}, L_v, T_1\)) and solve for the final equilibrium temperature \(T_f\): $$ 0.10 \cdot 4.18 \times 10^3 \cdot (T_f - 19) + 0.035 \cdot 0.897 \times 10^3 \cdot (T_f - 19) = 0.010 \cdot 2.26 \times 10^6 + 0.010 \cdot 4.18 \times 10^3 \cdot (100 - T_f) $$ Solving for \(T_f\), we obtain: $$ T_f \approx 63.84^{\circ} \mathrm{C} $$ Hence, the final equilibrium temperature of water after steam is added is approximately \(63.84^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

The \(R\) factor for housing insulation gives the thermal resistance in units of \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\). A good wall for harsh climates, corresponding to about 10.0 in of fiberglass, has \(R=40.0 \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\) a) Determine the thermal resistance in SI units. b) Find the heat flow per square meter through a wall that has insulation with an \(R\) factor of 40.0 , with an outside temperature of \(-22.0^{\circ} \mathrm{C}\) and an inside temperature of \(23.0^{\circ} \mathrm{C}\)

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Water is an excellent coolant as a result of its very high heat capacity. Calculate the amount of heat that is required to change the temperature of \(10.0 \mathrm{~kg}\) of water by \(10.0 \mathrm{~K}\). Now calculate the kinetic energy of a car with \(m=1.00 \cdot 10^{3} \mathrm{~kg}\) moving at a speed of \(27.0 \mathrm{~m} / \mathrm{s}(60.0 \mathrm{mph}) .\) Compare the two quantities.

Which of the following does not radiate heat? a) ice cube b) liquid nitrogen c) liquid helium d) a device at \(T=0.010 \mathrm{~K}\) e) all of the above f) none of the above

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.
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