Question

A piece of metal of mass 0.2 kg is transferred from a furnace to 0.15 kg of water at 0 k contained in a vessel of insulating material. The temperature of water rises to 30 k . Neglecting all leakages of heat, calculate the temperature of furnace . Given , specific heat capacity of the metal is 0.031 kcal kg -1 k-1

Answer 1

Let T is the temperature of the furnace

Finally water and metal has temprature = 30 K

Water temperature change from 0K to 30K and metal temerature change from T to 30K

Since all leakages of heat we can neglect

So Heat transfer from the metal to water = Heat gained by water to increase temperature

Heat transfer from metal to change its temprature from T to 30 K is given by Q=m × c × ΔT

m = 0.2 kg , c = Specific heat capacity = 0.031 kcal kg^-1k^-1 , ΔT = T - 30

Q = 0.2*0.031* (T - 30 ) kcal = 0.2*0.031* (T - 30 )*1000 cal = 6.2 * (T - 30 ) cal


The specific heat of water is 1 cal g^-1k^-1

Mass of water = 0.15 kg = 150 gram , ΔT = (30 - 0)K= 30K, c = 1 cal g^-1k^-1

Heat required to raise temerature of water from 0 K to 30 K is equal to Q1 = m × c × ΔT

Q1 = 150*1*30 = 4500 cal

Q = Q1

6.2 * (T - 30 ) cal = 4500 cal

(T - 30 ) = 725.81 K

T = 725.81 + 30 K

T = 755.81 K

Answer : The temperature of furnace = 755.81 K