Question

steam boiler made up of Steel with 900 kg the wire contains 200 kg of water assuming 70% of heat is delivered to the boiler and water how much heat is required to raise the temperature of whole of from 10 degree 200 degree heat capacity of steel is 0.1 ke and heat capacity of water is Banke​

Answer 1

Answer:

238 megaJoules

Explanation:

n[energy conversion efficiency]= 0.7 or 70%

m[water]= 900 kg=900 000 g

m[water]= 400 kg= 400 000 g

s[steel]=

4.186 J/[g*C]

t1= 20 C

t2= 100 C

dT=100-20= 80 C

Q[100%]=

Q[steel]+ Q[water]

Q[100%]= m[steel]*dT+m[water]*s[water]*dT

Q[100%]= 900 000 g*0.46 J/[g*C]*80 C+ 400 000 g*4.186 J/[g*C]

80 C= 167.1 MJ

Q[70%]= Q[100%]/0.7= 238.7 MJ

The right answer is

238 megaJoules.