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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

In one of your rigorous workout sessions, you lost \(150 \mathrm{~g}\) of water through evaporation. Assume that the amount of work done by your body was \(1.80 \cdot 10^{5} \mathrm{~J}\) and that the heat required to evaporate the water came from your body. a) Find the loss in internal energy of your body, assuming the latent heat of vaporization is \(2.42 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}\). b) Determine the minimum number of food calories that must be consumed to replace the internal energy lost (1 food calorie \(=4186\) J).

Why does tile feel so much colder to your feet after a bath than a bath rug? Why is this effect more striking when your feet are cold?

It has been proposed that global warming could be offset by dispersing large quantities of dust in the upper atmosphere. Why would this work, and how?

For a class demonstration, your physics instructor pours \(1.00 \mathrm{~kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) over \(4.00 \mathrm{~kg}\) of ice at \(0.00^{\circ} \mathrm{C}\) and allows the system to reach equilibrium. He is then going to measure the temperature of the system. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: \(L_{\text {ice }}=3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}\), \(L_{\text {steam }}=2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}, c_{\text {water }}=4186 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right) .\) You are asked to calculate the final equilibrium temperature of the system. What value do you find?

A material has mass density \(\rho,\) volume \(V\), and specific heat \(c .\) Which of the following is a correct expression for the heat exchange that occurs when the material's temperature changes by \(\Delta T\) in degrees Celsius? a) \((\rho c / V) \Delta T\) b) \((\rho c V)(\Delta T+273.15)\) c) \((\rho c V) / \Delta T\) d) \(\rho c V \Delta T\)
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