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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