Question

25. 10 g of ice is at 0 °C. Calculate the total heat required to convert it into steam at 100 °C
Given Lice = 3.35 × 10^ J kg!
Lsteam = 2.26 x 100 J kg-
Specific heat of water = 4200 J kg 'K-
-11-​

Answer 1

At 0℃ both ice and water exist. Here first ice will change to water but the temperature will be same i.e it will become water at 0℃. You may be wondering but this is due to latent heat which is hidden inside. We can find out heat required by this process :

H = m×Lf (where m is mass in kg and Lf is latent heat of fusion and is value is fixes and is 3.35 × 10^5 J / kg for ice.)

So the amount of heat required to change 10 gram of ice to water is :( 10g = 1/100kg)

.01 × 3.35×10^5=3350J. -…………Eq1

Now after this on heating water will reach hoti 100 ℃. So about of heat required to change temperature is given by:

H = mc∆T. (where m is mass in kg , c is specific heat and it's value is fixed . For water it is 4200J/kg ,for ice it is 2100 J/kg and for steam it is 2010J /kg. And ∆T is change in temperature required.)

Here we have to change the temperature of water ,so we will apply for water.Heat required:

H= .01× 4200× (100–0)=4200J .......Eq2

Now on heating further water will change to steam . And the formula to find is mLv where m is mass in kg , Lv is latent heat of vaporization and it's value is 22.6J/kg for water.)

H= .01×22.6×10^5.=22600J. ......Eq3

On adding Eq (1),(2),(3) we will get the total heat required.

3350J + 4200J + 22600 J = 30150 J.

You can change this to calorie as

1 calorie = 4.184 J.

So 30150 J= 30150 ×4.184 calorie = 126147.6

calorie.