Question

How much heat is needed to transform 500g of water at 50 °C into steam at 100 °C ?​

Answer 1

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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 1 gram of ice to water is :( 1g = 1/1000kg)

.001 × 3.35×10^5=335J. -(equation 1)

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= .001× 4200× (100–0)=420J (equation 2)

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 22.5J/kg for water.)

H= .001×22.5×10^5.=2250J. (equation 3)

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

335J + 420J + 2250 J = 3005 J.

You can change this to calorie as

1 calorie = 4.184 J.

So 3005 J= 3005 ×4.184 calorie =718.21 calorie.

Answer 2

Answer:

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 1 gram of ice to water is :( 1g = 1/1000kg)

.001 × 3.35×10^5=335J. -(equation 1)

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= .001× 4200× (100–0)=420J (equation 2)

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 22.5J/kg for water.)

H= .001×22.5×10^5.=2250J. (equation 3)

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

335J + 420J + 2250 J = 3005 J.

You can change this to calorie as

1 calorie = 4.184 J.

So 3005 J= 3005 ×4.184 calorie =718.21 calorie.