Question

Q3) How much heat energy would be given off in cooling water of 2.5 Kg from
85°C to 35°C ? (Answer = 125 kcal)​

Answer 1

Explanation:

To convert 2.5 kg of water at 25°C to steam at 100 °C, first we need to convert water at 25 °C to water at 100 °C and then water at 100 °C to steam at 100 °C.

For first step,

Q1= m*c*ΔΦ

Q1= 2500*1*75 cal

Q1=1,87,500 cal

For second step,

Q2= m*L

Q2= 2500*540 cal

Q2=13,50,000 cal

Thus total heat,

Q= Q1 + Q2

Q= 1,87,500 + 13,50,000

Q= 15,37,500 cal

Q= 1537.5 kcal

Q= 1537.5*4.184 kJ

Q= 6432.9 kJ

Answer 2

Answer:

since Q=mc((del )t) Q=25/10×1×(85-35) =2.5×50 =125cal.

I hope this helps you please mark this answer as brainalist and follow me