Problem Set 14: | 1 | 3 | 7 | 9 | 15 | 19 | 21 | 23 | 25 | 29 | Go up
 - by Elaine Fu, 2003

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

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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

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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

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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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For this problem we need two equations Ke=1/2mv² 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

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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_ironc_ironD T_iron= m_aluminumc_aluminumD T_aluminum + m_glycerinc_glycerinD T_glycerin solve for c_glycerin

c_glycerin=(m_ironc_ironD T_iron-m_aaluminumc_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

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.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 ^oC to 100.0 ^oC (A change of 92 ^oC), adding heat with the formula Q = mcDT

Once you know the heat added, that becomes the work that the heater does, and we will use the relationship between work, time and power to solve for time.

For the water from 8.0 ^oC to 100.0 ^oC: Q = (.60 kg)(4186 J/kg^oC)(92.0 ^oC) = 231,067.2 J
For the Al      from 8.0 ^oC to 100.0 ^oC: Q = (.360 kg)(900 J/kg^oC)(92.0 ^oC) = 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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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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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_iceL_{ice(solid to liquid)} + m_icec_{ice(-8.5° C to 0° C)D} T_{ice(-8.5° C to 0° C)} + m_icec_{ice(0° C to 17° C)D} T_{ice(0° C to 17° C)} = m_aluminumc_aluminumD T_aluminum+ m_waterc_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_aluminumc_aluminumD T_aluminum+ m_waterc_waterD T_water

then move everything to the right so that the m_ice is all by it self

m_ice = (m_aluminumc_aluminumD T_aluminum+ m_waterc_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

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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_HgL_Hg + m_Hgc_HgD T_{(-39.0° C to 5.06° C)} = m_waterc_waterD T + m_aluminumc_aluminumD T

solve for L_Hg First move everything to the right then divide the whole thing by mercury mass.

L_Hg = {[m_waterc_waterD T_water + m_aluminumc_aluminumD T_aluminum] - [m_Hgc_{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

1.12x10⁴ J/kg
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