When p calories of heat power of body will be is given to a body it gives q calories ,then the absorption
Answers
Answer:
When two bodies at different temperatures come in contact, heat is transferred from the body at a higher temperature to the body at a lower temperature. According to the principle of calorimetry (energy conservation), heat loss is equal to the heat gained. The heat gained by a body is related to the rise in its temperature
Δ
T
by
Q
=
m
s
Δ
T
where
m
is the mass of the body and
s
is the specific heat of the material of the body.
When there is a change of state (phase transitions like fusion and vaporization), the body gain or lose heat without a change in temperature. The amount of heat transfer is given by
Q
=
m
L
where
L
is latent heat of the material of the body. For water, the latent heat of fusion is 80 cal/g and the latent heat of vaporization is 540 cal/g.
Heat Added and Temperature Change
Added heat increases temperature of a body or changes its state (fusion, vaporization) at a constant temperature.
The work done on a body can be used to increase its temperature. The work (in J) is related to the heat (in cal) by
Q
=
W
J
, where
J
=
4.18
cal/g is called the mechanical equivalent of heat.
A calorimeter is used to measure the amount of heat transferred when two bodies are brought in contact. The water equivalent of a calorimeter is the mass of water that would need the same quantity of heat transfer as the calorimeter to create the same temperature change. Its unit is kg.
Mixing Two Substances Without Change of State: Consider mixing of two substances of masses
m
1
and
m
2
, specific heats
s
1
and
s
2
and temperatures
T
1
and
T
2
(
<
T
1
)
. Let temperature of the mixture be
T
. By principle of calorimetry,
m
1
s
1
(
T
1
−
T
)
=
m
2
s
2
(
T
−
T
2
)
.
Solve to get temperature of the mixture at equilibrium as
T
=
m
1
s
1
T
1
+
m
2
s
2
T
2
m
1
s
1
+
m
2
s
2
Mixing Two Substances With Change of State: Suppose water at temperature
T
w
deg C is mixed with ice at 0 deg C. In this case, first ice will melt and then its temperature will rise to attain thermal equilibrium. It is important to consider the latent heat during phase transition.
Solved Problems on Calorimetry
Problem from IIT JEE 2005
Water of volume 2 litre in a container is heated with a coil of 1 kW at
27
o
C
. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from
27
o
C
to
77
o
C
? (Specific heat of water is 4.2 kJ/kg.)
8 min 20 second
6 min 2 second
7 min
14 min
Solution: The heater coil gives energy at a rate of 1000 J/s, out of which 160 J/s is dissipated through the lid. Thus, energy for heating the water is available at a rate of,
P
=
1000
−
160
=
840
J
/
s
.
The heat energy required to raise the temperature of mass
m
of water from
T
1
to
T
2
is,
Q
=
m
S
(
T
2
−
T
1
)
,
where
S
is specific heat of the water. If
t
is the time required to raise temperature of mass
m
from
T
1
to
T
2
then, by energy conservation,
P
t
=
Q
.
The density of water is
1000
k
g
/
m
3
. Thus, mass of 2 litre of water is
m
=
2
k
g
. Substitute
P
and
Q
from first and second equations into third equation to get,
t
=
m
S
(
T
2
−
T
1
)
P
=
500
s
e
c
=
8
m
i
n
20
s
e
c
.
Problem from IIT JEE 2005
One calorie is defined as the amount of heat required to raise temperature of 1 g of water by
1
o
C
in a certain interval of temperature and at certain pressure. The temperature interval and pressure is,
from
14.5
o
C
to
15.5
o
C
at 760 mm of Hg
from
98.5
o
C
to
99.5
o
C
at 760 mm of Hg
from
13.5
o
C
to
14.5
o
C
at 76 mm of Hg
from
3.5
o
C
to
4.5
o
C
at 76 mm of Hg
Solution: One calorie is defined as the heat required to raise temperature of 1 g of water from
14.5
o
C
to
15.5
o
C
at atmospheric pressure.
Absorption of Heat Power
Explanation:
- A calorimeter is utilized to quantify the measure of warmth moved when two bodies are gotten contact. What might be compared to a calorimeter is the mass of water that would require a similar amount of warmth move as the calorimeter to make a similar temperature change. Its unit is kg
- Calories of incident heat=p
- Calories of absorbed heat=q
- ∵ we know that coefficient of absorption is the ratio of heat absorbed to the incident heat.
- So, coefficient of absorption=