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Chapter 14: Problem 33

What mass of water at \(25.0^{\circ} \mathrm{C}\) added to a Styrofoam cup containing two 50.0 -g ice cubes from a freezer at \(-15.0^{\circ} \mathrm{C}\) will result in a final temperature of \(5.0^{\circ} \mathrm{C}\) for the drink?

Short Answer

Expert verified
Answer: 1648 g

Step by step solution

01

Calculate the heat required to raise the temperature of ice cubes to 0°C

We first need to calculate the heat required to raise the temperature of both ice cubes (-15.0°C) to 0°C. We will use the equation \(Q = mcΔT\): \(Q = (m_{ice})(c_{ice})(ΔT_{ice})\) Here, \(m_{ice} = 2 \times 50.0\,\text{g} = 100.0\,\text{g}\) (mass of both ice cubes), \(c_{ice} = 2.093\,\frac{J}{g°C}\) (specific heat capacity of ice), and \(ΔT_{ice} = 15.0\,°\text{C}\) (change in temperature from -15.0°C to 0°C). \(Q_{ice} = (100.0\,\text{g})(2.093\,\frac{J}{g°C})(15.0\,°\text{C}) = 3140\,\text{J}\) (rounded to 3 significant figures)
02

Calculate the heat required to change ice from solid to liquid

Next, we need to calculate the heat required to change the solid ice cubes into liquid water at 0°C. We will use the equation \(Q = mL\): \(Q = (m_{ice})(L_{fusion})\) Here, \(m_{ice} = 100.0\,\text{g}\) (mass of both ice cubes) and \(L_{fusion} = 334\,\frac{J}{g}\) (latent heat of fusion for water). \(Q_{fusion} = (100.0\,\text{g})(334\,\frac{J}{g}) = 33400\,\text{J}\)
03

Calculate the heat required to raise the temperature of liquid water to the final temperature

Now, we need to calculate the heat required to raise the temperature of the entire mixture (100 g of liquid water and the mass of water) to the final temperature of 5.0°C. We will use the equation \(Q = mcΔT\): \(Q = (m_{water}+m_{ice})(c_{water})(ΔT_{water})\) Here, \(m_{water} + m_{ice} = 100.0\,\text{g} + m_{water}\) (mass of water added plus the mass of ice cubes), \(c_{water} = 4.18\,\frac{J}{g°C}\) (specific heat capacity of water), and \(ΔT_{water} = 5.0\,°\text{C}\) (change in temperature from 0°C to 5.0°C). \(Q_{water} = (100.0\,\text{g} + m_{water})(4.18\,\frac{J}{g°C})(5.0\,°\text{C}) = (100.0\,\text{g} + m_{water})\times 20.9\,\text{J} \)
04

Equate the heat lost by the water to the heat gained by the ice

The heat lost by the water (\(Q_{water}\)) will be equal to the heat gained by the ice (\(Q_{total}\)), where \(Q_{total} = Q_{ice} + Q_{fusion}\): \(Q_{water} = Q_{total}\) \((100.0\,\text{g} + m_{water})\times 20.9\,\text{J} = 3140\,\text{J} + 33400\,\text{J}\)
05

Solve for the mass of water

Now, we can solve for the mass of water (\(m_{water}\)) by simplifying the equation from Step 4: \((100.0\,\text{g} + m_{water})\times 20.9\,\text{J} = 36540\,\text{J}\) Divide both sides by 20.9 J: \(100.0\,\text{g} + m_{water} = 1748\,\text{g}\) (rounded to 3 significant figures) Subtract 100.0 g from both sides: \(m_{water} = 1648\,\text{g}\) Therefore, 1648 g of water at 25.0°C needs to be added to the two 50.0-g ice cubes to achieve a final temperature of 5.0°C for the drink.

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Most popular questions from this chapter

A 2.0 -kg block of copper at \(100.0^{\circ} \mathrm{C}\) is placed into $1.0 \mathrm{kg}$ of water in a 2.0 -kg iron pot. The water and the iron pot are at \(25.0^{\circ} \mathrm{C}\) just before the copper block is placed into the pot. What is the final temperature of the water, assuming negligible heat flow to the environment?
A metal rod with a diameter of \(2.30 \mathrm{cm}\) and length of $1.10 \mathrm{m}\( has one end immersed in ice at \)32.0^{\circ} \mathrm{F}$ and the other end in boiling water at \(212^{\circ} \mathrm{F}\). If the ice melts at a rate of 1.32 g every 175 s, what is the thermal conductivity of this metal? Identify the metal. Assume there is no heat lost to the surrounding air.
If \(125.6 \mathrm{kJ}\) of heat are supplied to \(5.00 \times 10^{2} \mathrm{g}\) of water at \(22^{\circ} \mathrm{C},\) what is the final temperature of the water?
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.]
A blacksmith heats a 0.38 -kg piece of iron to \(498^{\circ} \mathrm{C}\) in his forge. After shaping it into a decorative design, he places it into a bucket of water to cool. If the available water is at \(20.0^{\circ} \mathrm{C},\) what minimum amount of water must be in the bucket to cool the iron to \(23.0^{\circ} \mathrm{C} ?\) The water in the bucket should remain in the liquid phase.
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