FE Physics

Formulas & Explanation

Formulas & Explanation

Dutch version

**Heat**

**
**

Q = mc
ΔT

Q
amount of heat in J

m
mass in kg

c
specific heat capacity in Jkg^{-1o}C^{-1}
or Jkg^{-1}K^{-1}

ΔT
change in temperature in ^{o}C of K

The specific heat capacity of
water : 4.18 x 10^{3} Jkg^{-1}K^{-1}

*
Example 1*

500 g water is heated in a
pan. The temperature increases of 20^{o}C to 50 ^{o}C.

The heat capacity of the pan
is neglected.

Calculate the amount of
energy needed to heat the water.

Q = mc
ΔT** **

Q = (0.500)( 4.18 x 10^{3}
) (50-20) = 6.27 x
10^{4} J

(at a temperature drop of 30
^{o} C, an energy of
6.27 x 10^{4} J is released)

C : heat capacity : the gained/lost
energy per degree rise/drop in temperature

C = Q/ΔT

also applies : C = mc.

C
heat capacity
in JK^{-1 }or in J ^{
o}C^{-1}

*Example 2*

We start from the example
above, but now the pan in which the water is heated
has a heat capacity of C = 40 JK^{-1}

Q = mc
ΔT
+ C
ΔT
= 6.27 x 10^{4} + 40 (50-20) = 6.27 x 10^{4} + 0.12 x 10^{4}
= 6.39 x 10^{4 }J

**Efficiency
(η****)**

η = P_{useful}/P_{in}
or
η
= E_{useful}/E_{in }

_{
}

η
efficiency
(no dimension)

P_{useful } useful
power out in W

P_{in} total
power in in W

E_{useful }useful
energy out in J

E_{in }total
energy in in J

_{
}

*
Example 3*

In example 2 the water is
heated by a an electric heating element with a power of
1000 W.

The water was heated for
1.5 minutes.

Calculate the efficiency of
the process

E_{in }= P
t = (1000) ( 1.5 x 60)
= 9.0 x 10^{4}
J

E_{useful }= Q = 6.39 x 10^{4}
J

η
= (6.39 x 10^{4})/(9.0
x 10^{4})
= 0.71 or
71 %