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You leave a pastry in the refrigerator on a plate and ask your roommate to take it out before you get home so you can eat it at room temperature, the way you like it. Instead, your roommate plays video games for hours. When you return, you notice that the pastry is still cold, but the game console has become hot. Annoyed, and knowing that the pastry will not be good if it is microwaved, you warm up the pastry by unplugging the console and putting it in a clean trash bag (which acts as a perfect calorimeter) with the pastry on the plate. After a while, you find that the equilibrium temperature is a nice, warm $38.3\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ . You know that the game console has a mass of 2.1 kg. Approximate it as having a uniform initial temperature of $45\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ . The pastry has a mass of 0.16 kg and a specific heat of $3.0\phantom{\rule{0.2em}{0ex}}\text{k}\phantom{\rule{0.2em}{0ex}}\text{J/}(\text{kg}\xb7\text{\xbaC}),$ and is at a uniform initial temperature of $4.0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ . The plate is at the same temperature and has a mass of 0.24 kg and a specific heat of $0.90\phantom{\rule{0.2em}{0ex}}\text{J/}(\text{kg}\xb7\text{\xbaC})$ . What is the specific heat of the console?
$1.7\phantom{\rule{0.2em}{0ex}}\text{kJ/}(\text{kg}\xb7\text{\xbaC})$
Two solid spheres, A and B , made of the same material, are at temperatures of $0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ and $100\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ , respectively. The spheres are placed in thermal contact in an ideal calorimeter, and they reach an equilibrium temperature of $20\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ . Which is the bigger sphere? What is the ratio of their diameters?
In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of $808\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}.$ (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to $3.00\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ . (Use ${c}_{\text{P}}$ and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c) Compare the amount of cooling obtained from melting an identical mass of $0\text{-}\text{\xb0}\text{C}$ ice with that from evaporating the liquid nitrogen.
a. $1.57\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{kcal}$ ; b. $18.3\phantom{\rule{0.2em}{0ex}}\text{kW}\xb7\text{h}$ ; c. $1.29\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{kcal}$
Some gun fanciers make their own bullets, which involves melting lead and casting it into lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from $25.0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ ?
A 0.800-kg iron cylinder at a temperature of $1.00\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ is dropped into an insulated chest of 1.00 kg of ice at its melting point. What is the final temperature, and how much ice has melted?
$6.3\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ . All of the ice melted.
Repeat the preceding problem with 2.00 kg of ice instead of 1.00 kg.
Repeat the preceding problem with 0.500 kg of ice, assuming that the ice is initially in a copper container of mass 1.50 kg in equilibrium with the ice.
$63.9\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ , all the ice melted
A 30.0-g ice cube at its melting point is dropped into an aluminum calorimeter of mass 100.0 g in equilibrium at $24.0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ with 300.0 g of an unknown liquid. The final temperature is $4.0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}$ . What is the heat capacity of the liquid?
(a) Calculate the rate of heat conduction through a double-paned window that has a $1.50{\text{-m}}^{2}$ area and is made of two panes of 0.800-cm-thick glass separated by a 1.00-cm air gap. The inside surface temperature is $15.0\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C},$ while that on the outside is $\mathrm{-10.0}\phantom{\rule{0.2em}{0ex}}\text{\xb0}\text{C}.$ ( Hint: There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a).
a. 83 W; b. $1.97\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{W}$ ; The single-pane window has a rate of heat conduction equal to 1969/83, or 24 times that of a double-pane window.
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