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Chapter 14: Problem 43

You are given \(250 \mathrm{g}\) of coffee (same specific heat as water) at \(80.0^{\circ} \mathrm{C}\) (too hot to drink). In order to cool this to $\left.60.0^{\circ} \mathrm{C}, \text { how much ice (at } 0.0^{\circ} \mathrm{C}\right)$ must be added? Ignore heat content of the cup and heat exchanges with the surroundings.

Short Answer

Expert verified
Answer: Approximately \(35.7 \mathrm{g}\) of ice is needed to cool the coffee to \(60.0^{\circ} \mathrm{C}\).

Step by step solution

01

Write the energy conservation equation

The energy conservation equation can be written as: heat gained by the ice + heat gained by the melted ice = heat lost by the coffee
02

Write the formula for heat gained/lost

Heat gained/lost can be written as \(Q = mcΔT\) for a given substance, where \(Q\) is the heat, \(m\) is the mass, \(c\) is the specific heat capacity, and \(ΔT\) is the change in temperature. In this case, we have three substances: ice, melted ice (water), and coffee.
03

Introduce the latent heat of fusion for ice

During ice melting, we must account for the amount of heat absorbed by ice to turn to water. The heat obtained from the coffee will first be used to melt the ice; the melted ice will then gain heat from the remaining heat available. The latent heat of fusion can be written as \(Q_{fusion} = mL\), where \(Q_{fusion}\) is the heat absorbed during the phase change, \(m\) is the mass of ice, and \(L\) is the latent heat of fusion.
04

Write the energy conservation equation with known values and variables

Considering all the steps above, we can write the equation for this problem, taking \(m_{ice}\) as the mass of ice added: \((m_{ice}L) + (m_{ice}c_wΔT_{melted\_ice}) = (m_{coffee}c_wΔT_{coffee})\) where \(c_w = 4.18 \mathrm{J/g\:C}\) is the specific heat capacity of water, \(L = 334 \mathrm{J/g}\) is the latent heat of fusion for water, \(m_{coffee} = 250 \mathrm{g}\), \(ΔT_{melted\_ice} = (60 - 0)\), and \(ΔT_{coffee} = (80 - 60)\).
05

Solve the equation for mass of ice

Simplify and plug in known values: \(m_{ice}(334) + m_{ice}(4.18)(60) = (250)(4.18)(20)\) \(m_{ice}(334 + 250.8) = (250)(83.6)\) \(m_{ice} = \frac{(250)(83.6)}{584.8}\) \(m_{ice} \approx 35.7 \mathrm{g}\) So, approximately \(35.7 \mathrm{g}\) of ice is needed to cool the coffee to \(60.0^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

While camping, some students decide to make hot chocolate by heating water with a solar heater that focuses sunlight onto a small area. Sunlight falls on their solar heater, of area \(1.5 \mathrm{m}^{2},\) with an intensity of $750 \mathrm{W} / \mathrm{m}^{2}$ How long will it take 1.0 L of water at \(15.0^{\circ} \mathrm{C}\) to rise to a boiling temperature of $100.0^{\circ} \mathrm{C} ?$
Compute the heat of fusion of a substance from these data: \(31.15 \mathrm{kJ}\) will change \(0.500 \mathrm{kg}\) of the solid at \(21^{\circ} \mathrm{C}\) to liquid at \(327^{\circ} \mathrm{C},\) the melting point. The specific heat of the solid is \(0.129 \mathrm{kJ} /(\mathrm{kg} \cdot \mathrm{K})\).
A chamber with a fixed volume of \(1.0 \mathrm{m}^{3}\) contains a monatomic gas at \(3.00 \times 10^{2} \mathrm{K} .\) The chamber is heated to a temperature of \(4.00 \times 10^{2} \mathrm{K}\). This operation requires \(10.0 \mathrm{J}\) of heat. (Assume all the energy is transferred to the gas.) How many gas molecules are in the chamber?
Bare, dark-colored basalt has a thermal conductivity of 3.1 $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ whereas light-colored sandstone's thermal conductivity is only \(2.4 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) Even though the same amount of radiation is incident on both and their surface temperatures are the same, the temperature gradient within the two materials will differ. For the same patch of area, what is the ratio of the depth in basalt as compared with the depth in sandstone that gives the same temperature difference?
At a tea party, a coffeepot and a teapot are placed on the serving table. The coffeepot is a shiny silver-plated pot with emissivity of \(0.12 ;\) the teapot is ceramic and has an emissivity of \(0.65 .\) Both pots hold \(1.00 \mathrm{L}\) of liquid at \(98^{\circ} \mathrm{C}\) when the party begins. If the room temperature is at \(25^{\circ} \mathrm{C},\) what is the rate of radiative heat loss from the two pots? [Hint: To find the surface area, approximate the pots with cubes of similar volume.]
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