Question

Find the amount of heat in calories needed to heat 3.20 mol of liquid water from 25◦C to 95◦C.​

Answer 1

Answer:

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