Answer to Question #218623 in Molecular Physics | Thermodynamics for Pandrake

Question

 2.5g of water at 100K occupies a volume of 2.5cm3 at atmospheric pressure. When the water is boiled, it occupies

4178cm3 as steam. What is the change in its internal energy? The latent heat of vaporization is 2.26106 J / kg. the pressure of one atm is 1.013105 Nm.2 

Answer 1

Given:-

Mass (m)= 2.5 g = 0.0025 Kg

Temp. of water (T₁)= 100 K = 0°C

Specific heat Capacity (C_p)= 4.186 joule/gram °C

Latent heat of water(L)= 2230 joules/gram

Final Temp (T₂)= 100°C

To Find:-

Change in Internal Energy = mC_p$\Delta T + mL$

= 2.5*4.186*100 + 2.5*2230

= 6621.5 J