Question

Exactly 10.0 mL of water at 25.0°C is added to a hot iron skillet. All the water is converted into steam at 100.0°C. The mass of the pan is 20 kg. What is the change in temperature of the skillet? (Specific heat capacity of iron is 0.444 J/g°C)

Answer 1

For, water the temperature first changed from 25^oC to 100^oC, i..e. BOILING POINT OF WATER

Now, mass of water = volume of water = 10 g (considering the density of water = 1 g/ml)

specific heat of water = 4.2 J/g°C

temperature change = 25^oC - 100^oC = 75^oC

So, heat absorbed (Q₁) = mass x specific heat x temperature change = 10 g x 4.2 J/g°C x 75 °C = 3150 J

Next, water takes up latent heat and gets vaporised at the same temperature

Now, latent heat of vaporization of water = 2264.7 J/g

So, heat absorbed (Q₂) = mass x latent heat = 10 gm x 2264.7 J/g = 22647 J

So, total heat absorbed by water = (Q₁ + Q₂) = (3150 + 22647) J = 25797 J

So, the same heat is lost from the iron pan.

Now, mass of iron pan = 20 Kg = 20000 g

and specific heat of iron = 0.444 J/g°C

So heat lost = mass x specefic heat x temperature change

Or, 25797 J = 20000 g x 0.444 J/g°C x temperature change

So, temperature change = 2.905 °C

So, temperature change of the skillet = 2.905 °C