In order to raise the temperature of 100 grams of water from room temperature (20oC) to boiling temperature (100oC), an energy input of 33,440.0 J is required. If one uses a microwave oven (, of radiation 1.20 cm) to achieve this, how many photons of the microwave radiation are required?
Answers
Answer:
Explanation:
or starters, you need to know the mass of water present in your sample. Since no mention of what to use for the density of water, you can assume it to be equal to
1 g mL
−
1
.
This means that your sample contains
145
mL
⋅
1 g
1
mL
=
145 g
Now, you know that the amount of heat needed to heat the water from
21.0
∘
C
to liquid water at
100.0
∘
C
can be calculated using the equation
q
=
m
⋅
c
⋅
Δ
T
−−−−−−−−−−−−−
Here
q
is the heat lost or gained by the water
m
is the mass of the water
c
is the specific heat of water,, equal to
4.18 J g
−
1
∘
C
−
1
Δ
T
is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample
In your case, you have
Δ
T
=
100.0
∘
C
−
21.0
∘
C
=
79.0
∘
C
which means that the heat needed for this sample is
q
=
145
g
⋅
4.18 J
g
−
1
∘
C
−
1
⋅
79.0
∘
C
q
=
47,882 J