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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掳C

We first need to calculate the heat required to raise the temperature of both ice cubes (-15.0掳C) to 0掳C. We will use the equation \(Q = mc螖T\): \(Q = (m_{ice})(c_{ice})(螖T_{ice})\) Here, \(m_{ice} = 2 \times 50.0\,\text{g} = 100.0\,\text{g}\) (mass of both ice cubes), \(c_{ice} = 2.093\,\frac{J}{g掳C}\) (specific heat capacity of ice), and \(螖T_{ice} = 15.0\,掳\text{C}\) (change in temperature from -15.0掳C to 0掳C). \(Q_{ice} = (100.0\,\text{g})(2.093\,\frac{J}{g掳C})(15.0\,掳\text{C}) = 3140\,\text{J}\) (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掳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掳C. We will use the equation \(Q = mc螖T\): \(Q = (m_{water}+m_{ice})(c_{water})(螖T_{water})\) Here, \(m_{water} + m_{ice} = 100.0\,\text{g} + m_{water}\) (mass of water added plus the mass of ice cubes), \(c_{water} = 4.18\,\frac{J}{g掳C}\) (specific heat capacity of water), and \(螖T_{water} = 5.0\,掳\text{C}\) (change in temperature from 0掳C to 5.0掳C). \(Q_{water} = (100.0\,\text{g} + m_{water})(4.18\,\frac{J}{g掳C})(5.0\,掳\text{C}) = (100.0\,\text{g} + m_{water})\times 20.9\,\text{J} \)
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掳C needs to be added to the two 50.0-g ice cubes to achieve a final temperature of 5.0掳C for the drink.

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Most popular questions from this chapter

Consider the leaf of Problem \(70 .\) Assume that the top surface of the leaf absorbs \(70.0 \%\) of \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2}\) of radiant energy, while the bottom surface absorbs all of the radiant energy incident on it due to its surroundings at \(25.0^{\circ} \mathrm{C} .\) (a) If the only method of heat loss for the leaf were thermal radiation, what would be the temperature of the leaf? (Assume that the leaf radiates like a blackbody.) (b) If the leaf is to remain at a temperature of \(25.0^{\circ} \mathrm{C}\) how much power per unit area must be lost by other methods such as transpiration (evaporative heat loss)?
It takes \(880 \mathrm{J}\) to raise the temperature of \(350 \mathrm{g}\) of lead from 0 to \(20.0^{\circ} \mathrm{C} .\) What is the specific heat of lead?
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 10.0 -g iron bullet with a speed of $4.00 \times 10^{2} \mathrm{m} / \mathrm{s}\( and a temperature of \)20.0^{\circ} \mathrm{C}$ is stopped in a \(0.500-\mathrm{kg}\) block of wood, also at \(20.0^{\circ} \mathrm{C} .\) (a) At first all of the bullet's kinetic energy goes into the internal energy of the bullet. Calculate the temperature increase of the bullet. (b) After a short time the bullet and the block come to the same temperature \(T\). Calculate \(T\), assuming no heat is lost to the environment.
A hiker is wearing wool clothing of \(0.50-\mathrm{cm}\) thickness to keep warm. Her skin temperature is \(35^{\circ} \mathrm{C}\) and the outside temperature is \(4.0^{\circ} \mathrm{C} .\) Her body surface area is \(1.2 \mathrm{m}^{2} .\) (a) If the thermal conductivity of wool is $0.040 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ what is the rate of heat conduction through her clothing? (b) If the hiker is caught in a rainstorm, the thermal conductivity of the soaked wool increases to \(0.60 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) (that of water). Now what is the rate of heat conduction?
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