1.
A pot is filled with 5 kg of water at 20oC.
Calculate how much heat energy would be needed to raise the temperature to 70oC.
[specific heat capacity of water = 4200J/kgoC ]
Answer :
Given : m = 5 kg
Cwater
= 4200J/kg0C
ΔT = 700C
– 200C = 500C
Find : q = ….?
Formula : q = m c ΔT
= 5 kg x 4200J/kgoC x 50oC
= 1050000 J
= 1.05 MJ
2.
How many joules of heat are needed to
raise the temperature of 10.0 g of aluminum from 24°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C
Answer
:
Given : m = 10 g = 0.01 kg
Caluminum
= 0.90 J/kg0C
ΔT = 55oC
– 240C = 31oC
Find : q = ….?
Formula : q = m c ΔT
= 0.01 kg
x 0.90 J/kgoC x 31oC
= 0.279 J
3.
Calculate the heat capacity of a piece of
wood if 300.0 g of the wood absorbs 1.35×104 Joules of heat and its
temperature changes from 24°C to 49°C.
Answer
:
Given : m = 300 g = 0.3 kg
q= 1.35 x 104 J
ΔT = 49oC
– 240C = 25oC
Find : c = ….?
Formula : c = q : ( m ΔT )
= 1.35 x 104J
: ( 0.3 kg x 25oC )
= 1.8 x 103
J/kgoC
4.
The temperature of 250 g of water changed
from 24oC to 52oC. How much heat did this sample absorb?
Cwater = 4.18 J/goC
Answer
:
Given : m = 250 g
Cwater
= 4.18 J/g0C
ΔT = 52oC
– 240C = 28oC
Find : q = ….?
Formula : q = m c ΔT
= 250g x
4.18 J/goC x 28oC
= 29260 J
= 29.26 kJ
5.
What is the specific heat of silicon if it
takes 576 J to raise the temperature of 45.0g of Silicon by 18.0oC?
Answer
:
Given : m = 45.0 g
q= 576 J
ΔT = 18oC
Find : c = ….?
Formula : c = q : ( m ΔT )
= 576J : (
45 g x 18oC)
= 0.71 J/goC
6.
What is the specific heat of lead if it
takes 32 J to raise the temperature of a 25g block by 10oC?
Answer
:
Given : m = 25 g
q= 32 J
ΔT = 10oC
Find : c = ….?
Formula : c = q : ( m ΔT )
= 32 J : (
25 g x 10oC )
= 0.128 J/goC
7.
250.0g of 22°C water is heated until its
temperature is 37°C. If the specific
heat of water is 4.18 J/g°C, calculate the amount of heat energy needed to
cause this rise in temperature.
Answer
:
Given : m = 250 g
Cwater
= 4.18 J/g0C
ΔT = 37oC
– 220C = 15oC
Find : q = ….?
Formula : q = m c ΔT
= 250g x
4.18J/goCx15oC
= 15 675 J
8.
A 0.25 g piece of copper is heated and
fashioned into a bracelet. The amount of energy transferred by heat to the
copper is 1950 J. If the specific heat of copper is 390 J/goC, what Is the
change of the copper's temperature?
Answer:
Given : m = 0.25 g
Ccopper
= 390 J/kg0C
q = 1950J
Find : ΔT = ….?
Formula : ΔT = q: (m c )
= 1950J : ( 0.25 g x
390J/goC )
= 20oC
