- Submitted By: Kelly-Heward
- Date Submitted: 05/22/2014 4:05 AM
- Category: Science
- Words: 1478
- Page: 6

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=1/2mv2 and D Q=mcD T. When the...