A frying pan of mass 2.5kg
Q) A frying pan of mass 2.5kg has a temperature of 40 degree Celsius. The specific heat capacity of the frying pan is 1000J/kg degree Celsius, what will be its final temperature when 4500J of heat energy is supplied to it?
Solution:
Given , Mass
of frying pan(m) = 2.5kg
Initial temperature of pan(t) = 40
degree Celsius
Specific heat capacity of pan(s) =
1000J/kg degree Celsius
Quantity of heat supplied(Q) = 4500
J
Final temperature of pan(T) = ?
We know that
,
Q = m x s x (T - t )
or, 4500
= 2.5 x 1000 x(T - 40)
or, T - 40 = 1.8
or, T = 40 + 1.8 = 41.8 degree Celsius
Hence, the
final temperature of the pan is 41.8 degree Celsius.
Q) A heater
of power 2000 W gives some heat energy for 1 hour, calculate the temperature
raised by that heat in 100 kg of water at 30 degree Celsius. Also calculate the
final temperature raised by that heat in 100 kg of water at 30 degree Celsius.
Also calculate the final temperature of the water[ Sp. heat capacity of water =
4200 J/kg degree Celsius]
Ans :
Here , Given
Power of
heater(P) = 2000 W
Time (T) = 1
hour = 60 min = 60 x 60 sec = 3600 sec
We know
that, energy (Q) = Power x Time
= 2000 x 3600
= 7200000 J
Mass of water (m) = 100 kg
Temperature of water (t) = 30 degree
Celsius
Final temperature of water(T) = ?
We have,
Q = m x s (T - t )
or,
T - t = Q/ms = 7200000/420000 = 17 degree Celsius( approx)
Therefore,
The temperature raised is 17 degree Celsius.
Again, T =
17 + t = 17 + 30 = 47 degree Celsius.
Q) Calculate
the amount of heat required to raise the temperature of 50 kg of water from 10
degree to 60 degree Celsius.( specific heat capacity of water is 4200 J/kg
degree Celsius)
Solution:
Here, Mass
of water (m) = 50 kg
Change in
temperature (dt) = 60 - 10 = 50 degree Celsius
Specific
heat capacity of water(s) = 4200 J/kg degree Celsius
Heat
required (Q) = ?
According to
the formula,
Q = m.s.dt = 50 x 4200 x 50 = 10500000
J
Therefore, Q
= 10500 kJ
Therefore ,10500
kJ of heat is required to raise the temperature of 50 kg of water from 10
degree to 60 degree Celsius.
Q) An
aluminum kettle of mass 300 gm has water of 1 kg at 15 degree Celsius. Water is
to be heated to 100 degree Celsius .What heat energy must be supplied?(
specific heat capacity of water is 4200 J/kg degree Celsius, specific heat
capacity of aluminum is 899 J/kg degree Celsius.
Solution:
Here, mass of
water(m) = 1 kg
Specific heat capacity of water(s) =
4200 J/kg degree Celsius
Mass of kettle(m') = 300gm = 0.3 kg
Specific heat capacity of
aluminum(s') = 899 J/kg degree Celsius
Initial temperature (t) = 15 degree
Celsius
Final temperature (T) = 100 degree
Celsius
For water,
Heat
absorbed by water (Q) = ms( T - t )
= 1 x 4200 x (100 - 15 )
= 4200 x 85
= 357000 J
For the
kettle,
Heat
absorbed by kettle (Q') = m' x s' x (T - t)
= 0.3 x 899 ( 100 - 15 )
= 0.3 x 899 x 85
= 22924.5 J
Therefore,
The total heat to be supplied Q'' = Q + Q'
= 357000 J + 22924.5 J
= 379924.5 J
Q) A ball of
copper weighing 400 gm is transferred from a furnace to 1 kg of water at 20
degree Celsius. The temperature of water rises to 50 degree Celsius. What is
the original temperature of the ball?
(Specific
heat capacity of copper is 400 J/kg degree Celsius and specific heat capacity
of water is 4200 J/kg degree Celsius)
Solution:
Mass of
copper (m) = 400 gm = 0.4 kg
Mass of
water (m') = 1 kg
Temperature
of water (t) = 20 degree Celsius
Final
temperature of water (t') = 50 degree Celsius
So,
Heat gained by water Q = m x s x dt
= 1 x 4200 x (50 - 20)
= 126000J
Again , heat
lost by copper = 0.4 x 400 x (t - 50)
From the
principle of Calorimetry,
Total heat lost by body = Total heat gained by body
or, 0.4 x 400(t - 50) = 126000
or,
t - 50 = 787.5
or, t = 837.5
Hence the
temperature of the copper ball is 837.5 degree Celsius.
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