Test: Calorimetry - JEE MCQ

Water is used as a coolant in automobiles radiators because it has high specific heat capacity. So, it absorbs a large amount of heat for a degree rise in temperature.

For the composite system, energy conservation yields no net energy flow in or out of the system.
Let final temperature be T
Then, heat absorbed by A+heat absorbed by B+heat absorbed by C=0

Here in the question m₁=m₂=m₃

Mass of Iron = 100g
Water Eq of caloriemeter = 10g
Mass of water = 240g
Let the Temp. of surface = 0^ºC
S_iron = 470J/kg°C 

Total heat gained = Total heat lost.

So,100/1000× 470 × (θ – 60) = 250/1000 × 4200 × (60 – 20)
⇒ 47θ – 47 × 60 = 25 × 42 × 40
⇒  θ = 4200 + 2820/47= 44820/47 =953.61°C

A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.

At constant pressure
dQ= nCpdT
=1*(5R/2)*2
=5R

Heat absorbed by 5g ice when it converted to at 0° C = 5 x 80 = 400 cal.
Heat liberated by 10g water at 10° C to 0° C = 100 cal
Hence there is 15g water at 0° C and 300 cal needs to be liberated , thus for some amount of water converts into ice, hence the temp of mixture is 0° C.

∆Q =mS∆T

A principle of calorimetry states that if there is no loss of heat in surrounding the total heat lost by hot body equal to the total heat gained by a cold body.
i.e. heat loss = heat gain