Problem Set 14: | 1 | 3 | 7 | 9 | 15 | 19 | 21 | 23 | 25 | 29 | Go up

** - by Elaine Fu, 2003**

1. How much heat (joules) is required to raise the temperature of 20.0 kg of water from 15° C to 95° C?

D Q=mcD T

m=20.0 kg

c=4186 J/kg° C

D T=95° C-15° C=80° C

D Q=(20.0 kg)(4186 J/kg° C)(80° C)=6697600 J

6.7x10^6 J

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3. To what temperature will 7700 J of work raise 3.0 kg of water initially at 10.0° C?

So we start out with the same equation as above D Q=mcD T but we change D T to (T2-T1) with T1 as initial temperature (10.0° C) and T2 as final temperature which is what we wanna find out. D T is changed because we don't know the change in temperature we only know what it started at. Now the equation looks like this D Q=mc(T2-T1).

D Q=7700J

m=3.0 kg

T1=10.0° C

T2= D Q/mc+T1

T2=(7700 J)/(3.0 kg * 4186 J/kg° C)+10.0° C=10.613154961° C

10.6° C

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7. A water heater can generate 7200 kcal/h. How much water can it heat from 15° C to 50° C per hour?

The ratio is that 1 Cal (kcal) of heat can heat 1 kg of water by 1° C. Using this ratio, we divide the Q by the temperature change. You don’t need to calculate the time because it's per hour meaning we are only trying to figure out one hour here.

Q=7200 kcal/h

T2=50° C

T1=15° C

(7200 kcal/h)/(50° C-15° C)=205.7 kg/h

2.1x10^2 kg/h

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9. What is the specific hear of a metal substance if 135 kJ of heat is needed to raise 5.1 kg of the metal from 20° C to 30° C?

Going back to D Q=mcD T but solve for c.

c=D Q/(mD T)

D Q=135,000 J (it's increased by three zeros cos the problem has it as kJ)

m=5.1 kg

D T=30° C-20° C=10° C

c=(135,000 J)/(5.1 kg * 10° C)=2647.058 J/kg° C

2.6x10^3 J/kg° C

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15. The 1.20-kg head of a hammer has a speed of 8.0 m/s just before it strikes a nail and is brought to rest. Estimate the temperature rise of a 14-g iron nail generated by ten such hammer blows done in quick succession. Assume the nail absorbs all the energy.

For this problem we need two equations Ke=1/2mv^{2}and D Q=mcD T. When the hammer hits the nail the energy is turned into heat and transferred into the nail. So Ke=D Q meaning 1/2mv^2= mcD T. Solve for D T.

D T =(1/2mv^2)/(mc)

mass of hammer=1.20 kg

v=8.0 m/s

mass of nail=0.014kg

specific heat of iron=450 J/kg°C

D T=(1/2 * 1.20 kg * (8.0 m/s)^2) / (0.014 kg * 450 J/kg°C)=6.09523809524°C

But we are not done yet cos it's 10 blows to the nail not one. 6.09523809524°C * 10=60.9523809524°C

61°C

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19. When a 290-g piece of iron at 180° C is placed in a 100-g aluminum calorimeter cup containing 250 g of glycerin at 10° C, the final temperature is observed to be 38° C. What is the specific heat of glycerin?

This problem uses conservation of heat. Heat lost equals heat gained so eventhough iron looses heat, all of it is absorbed by the aluminum cup and the glycerin. From that we get iron=aluminim cup + glycerin

D Q_{iron}=D Q_{aluminum}+D Q_{glycerin}

m_{iron}c_{ironD }T_{iron}= m_{aluminum}c_{aluminumD }T_{aluminum }+ m_{glycerin}c_{glycerinD }T_{glycerin }solve for c_{glycerin}

c_{glycerin}=(m_{iron}c_{ironD }T_{iron}-m_{aaluminum}c_{aluminumD }T_{alunimun})/(m_{glycerin}T_{glycerin})

m_{iron}=0.290 kg

c_{iron}=450 J/kg° C

D T_{Iron}=180° C-38° C=142° C because the iron started at 180° C and ended at 38° C

m_{aluminum}=0.1 kg

c_{aluminum}=900 J/kg° C

D T_{aluminum}=38° C-10° C=28° C

m_{glycerin}=0.25 kg

D T_{glycerin}=38° C-10° C=28° C (the aluminum and glycerin have the same temperature cos the alumimun cup contains the glycerin so they are looked at as one element but their D Q are calculated separately)

c_{glycerin}= [(0.290 kg)(450 J/kg°C)(142°C)-(0.1 kg)(900 J/kg°C)(28°C)] / [(0.25 kg)(28°C)] = 2287.285 J/kg°C

2.3x10^3 J/kg°C

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21. How long does it take a 750-W coffeepot to bring to a boil 0.60 L of water initially at 8.0°C? Assume that the part of the pot which is heated with the water is made of 360 g of aluminum, and that no water boils away.

.6 liters of water has a mass of .60 kg, and 360 g is .360 kg of Al

To figure out the heat needed bring the water and aluminum from 8.0^{o}C to 100.0^{o}C (Achangeof 92^{o}C), adding heat with the formula Q = mcDTOnce you know the heat added, that becomes the

workthat the heater does, and we will use the relationship between work, time and power to solve for time.For the water from 8.0

^{o}C to 100.0^{o}C: Q = (.60 kg)(4186 J/kg^{o}C)(92.0^{o}C) = 231,067.2 J

For the Al from 8.0^{o}C to 100.0^{o}C: Q = (.360 kg)(900 J/kg^{o}C)(92.0^{o}C) = 29,808 J

So the net heat added is the sum of these amounts = 231,067.2 J + 29,808 J = 260875.2 J

Since Power = Work/Time, then Time = Work/Power = (260875.2 J)/(750 W) = 347.8336 s which works out to about 5 minutes and 48 seconds

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23. During exercise, a person may give off 180 kcal of heat in 30 min by evaporation of water from the skin. How much water has been lost?

Moving on to latent heat D Q = mL. Just plug in the numbers. L is given in the book on pg 425. Solve for m.

m = D Q/L

D Q = 180 Cal

L = 539 Cal/kg

m = (180 Cal) / (539 Cal/kg) = 0.333951762523 kg

0.334 kg of 0.334 liters

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25. A cube of ice is taken from the freezer at -8.5° C and placed in a 100-g aluminum calorimeter filled with 300 g of water at room temperature of 20° C. The final situation is observed to be all water at 17° C. What was the mass of the ice cube?

We have to look at this problem in steps. It's another conservation of heat problem but this time it also uses the fusion concept. There's the heat lost by the aluminum and the water and this loss equals heat gained by the ice. The tricky part is that the ice goes from -8.5° C to 17° C in three stages. It does the fusion thing first where it goes from a solid to liquid, then goes from -8.5° C up to 0° C, then finally from 0° C to 17° C. All of this translates to

Q_{ice }= Q_{aluminum }+ Q_{water }which then translates to

m_{ice}L_{ice(solid to liquid) }+ m_{ice}c_{ice(-8.5° C to 0° C)D }T_{ice(-8.5° C to 0° C)}+ m_{ice}c_{ice(0° C to 17° C)D }T_{ice(0° C to 17° C) }= m_{aluminum}c_{aluminumD }T_{aluminum}+ m_{water}c_{waterD }T_{water}solve for m_{ice}.

We have to factor out the m_{ice}

m_{ice }(L_{ice(solid to liquid) }+ c_{ice(-8.5° C to 0° C)D }T_{ice(-8.5° C to 0° C)}+ c_{ice(0° C to 17° C)D }T_{ice(0° C to 17° C}) = m_{aluminum}c_{aluminumD }T_{aluminum}+ m_{water}c_{waterD }T_{water}

then move everything to the right so that the m_{ice}is all by it self

m_{ice }= (m_{aluminum}c_{aluminumD }T_{aluminum}+ m_{water}c_{waterD }T_{water}) / (L_{ice(solid to liquid) }+ c_{ice(-8.5° C to 0° C)D }T_{ice(-8.5° C to 0° C)}+ c_{ice(0° C to 17° C)D }T_{ice(0° C to 17° C)})

m_{aluminum }= 0.1 kg

c_{aluminum }= 900 J/kg° C

D T_{aluminum}= 20° C - 17° C = 3° C

m_{water}= 0.3 kg

c_{water }= 4186 J/kg° C

D T_{water}= 20° C - 17° C = 3° C_{ }

L_{ice(solid to liquid)}= 3.33 x 10^5 J/kg

c_{ice(-8.5° C to 0° C) }= 2100 J/kg° C (it's not 4186 J/kg° C cos at this time it's still a ice solid not a liquid yet)

D T_{ice(-8.5° C to 0° C) }= 0° C -(-8.5° C) = 8.5° C

c_{ice(0° C to 17° C)}= 4186 J/kg° C

D T_{ice(0° C to 17° C }= 17° C - 0° C = 17° C

m_{ice}= (0.1 kg * 900 J/kg° C * 3° C + 0.3 kg * 4186 J/kg° C * 3° C) / (3.33 x 10^5 J/kg + 2100 J/kg° C * 8.5° C + 4186 J/kg° C * 17° C)

= 0.00956702653 kg

9.6 g

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29. The specific heat of mercury is 138 J/kg° C. Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of -39.0° C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of water at 12.80° C; the resulting equilibrium temperature is 5.06° C.

Just like the other problems heat gained equals heat lost. Heat gained by mercury is equivalent to heat lost by by the aluminum and water. So first the mercury has to go through the fusion step and dissolve into liquid, then the temperature goes from -39.0° C to 5.06° C.

m_{Hg}L_{Hg}+ m_{Hg}c_{HgD }T_{(-39.0° C to 5.06° C)}= m_{water}c_{waterD }T + m_{aluminum}c_{aluminumD }T

solve for L_{Hg }First move everything to the right then divide the whole thing by mercury mass.

L_{Hg }= {[m_{water}c_{waterD }T_{water}+ m_{aluminum}c_{aluminumD }T_{aluminum}] - [m_{Hg}c_{Hg(-39.0° C to 5.06° C)D }T_{(-39.0° C to 5.06° C)}]} / m_{Hg}

m_{water}= 0.400 kg

c_{water}= 4186 J/kg° C

D T_{water }= 12.8° C - 5.06° C = 7.74° C

m_{aluminum}= 0.620 kg

c_{aluminum}= 900 J/kg° C

D T_{aluminum }= 12.8° C - 5.06° C = 7.74° C

m_{Hg}= 1 kg

c_{Hg(-39.0° C to 0° C) }= 138 J/kg° C

D T_{(-39.0° C to 5.06° C) }= 5.06° C - (-39.0° C) = 44.06° C

L_{Hg}= [(0.400 kg * 4186 J/kg° C * 7.74° C + 0.620 kg * 900 J/kg° C * 7.74° C) - (1 kg * 138 J/kg° C * 44.06° C)] / 1 kg = 11896.776 J/kg

1.12x10^{4}J/kg

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