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Heat Transfer

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Q.

Two cylindrical rods of lengths l1 and l2, radii r1 and r2 have thermal conductivities k1 and k2 respectively. The ends of the rods are maintained at the same temperature difference. If l1 = 2l2 and r1 = r2/2, the rates of heat flow in them will be the same if k1/k2 is

Q.

If the temperature of a gas is increased from 27oC to 927oC, the root mean square speed of its molecules

Q.

At what temperature will oxygen molecules have the same root mean square speed as hydrogen molecules at 200 K?

Q.

743 J of heat energy is needed to raise the temperature of 5 moles of an ideal gas by 2 K at constant pressure. How much heat energy is needed to raise the temperature of the same mass of the gas by 2k at constant volume? Given, molar gas constant R = 8.3 J K – 1 mol – 1.

Q. 100 g of ice at 0C is mixed with 100 g of water 80C. The final temperature of the mixture will be?

The amount of heat required to convert 100 g of ice at 0C into water at 0C = 100 X 80 = 8000 calories. This is precisely the amount of heat lost by 100 g of water at 80C to bring its temperature down to 0C. Therefore, the temperature of the mixture remains 0C.

Q. two carnot engine A and B are operated in series. Engine receives heat from a reservoir at 600K and reject heat to a reservoir at temperature T.Engine B receives heat by engineA and in turn rejects it to a reservoir at 100K. If the efficiences of the two engines A and B are represented by n1 and n2 respectively. What is the value of n1/n2?

apply the efficiency formula twice , you will get the result

Q. A room at 20degree C spaceis heated by a heater of resistance 20 ohm connected to 200V mains. The temp. is uniform throughout the room and the heat is transmitted through a glass window of area 1 m2 and thickness 0.2cm. Calculate the temperature outside. Thermal conductivity of glass is 0.2 cal/mdegree Cs and mechanical equivalent of heat is 42. J/cal?

find the heat flux from the heater using V2/R formula , then equate this to k*A*dt/dx , here dt is difference in temperature and dx is 0.2cm

Q. 300 g of water at 25oC is added to 100 g of ice at 0oC. The final temperature of the mixture is?

Let the temperature of the mixture be toC. Heat lost by water in calories

= 300 X 1 X (25 - t)
= 7500 - 300t

Heat in calories required to melt 100 g of ice
= 100 X 80 = 8000


Now Heat lost = Heat gained

Or 7500 – 300t = 8000

Or t=-(5/3) C

Since t is negative, the water at 25C cools to 0C and melts a part of ice at 0C.

=> Heat lost = 300 X 1 X (25 - 0)
=> 7500 cal.

Hence only a part of the ice melts and resulting temperature is 0oC.

Q. The water equivalent of a copper calorimeter is 4.5 g. If the specific heat of copper is 0.09 cal.g^(–1o)C^(–1)?

(A)

m*s =4.5

m* (0.09) =4.5

m= 4.5 /0.09 =0.5 kg

(B)

water equivalent = (m*s) /Sw

Heat capacity = 4.5*1= 4.5 cal /º C

(C) Q= M*S * DELTA (T)

= 4.5*1.*8 = 36 cal

Since the temperature remains constant during the process of melting, no heat is exchanged with the calorimeter and hence

Q = 15 * 80 = 1200 cal.