In a heattreating process a 15kg metal part initially at 107

In a heat-treating process, a 1.5-kg metal part, initially at 1076 K, is quenched in a closed tank containing 95 kg of water, initially at 299K. There is negligible heat transfer between the contents of the tank and their surroundings. Modeling the metal part and water as incompressible with constant specific heats 0.5 kJ/kg-K and 4.4 kJ/kg-K, respectively, determine the final equilibrium temperature after quenching, in K.

Solution

For Metal Part:

m1=1.5 kg , T1=1076K , specific heat c1=0.5 kj/kg.k

for water

m2=95 kg ,T2= 299k ,c2=4.4 kj/kg.k

let final equllibrium temperature is T

Now heat realesed by metal part= heat absorbed by water ( assumption is no other heat loss)

m1*c1*(T1-T)=m2*c2*(T-T2)

1.5 kg* 0.5 KJ/kg.K*(1076-T)=95kg*4.4KJ/kg.K*(T-299)

solve for T

T=300.393 K