# Student

## Student

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
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
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
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