Question

Part A Determine the work done by 4.00 kg of water when it is all boiled to steam at 100 C. Assume a constant pressure of 1.00 atm. W- Submit Request Answer Part B Determine the change in internal energy of 4.00 kg of water when it is all boiled to steam at 100 °C. Assume a constant pressure of 1.00 atm Submit Request Answer

Answer 1

Given the mass of water is = 4 k.g

We know the density of water is = 1000 k.g/m³

since density = mass / volume

So volume of the given mass of water is = V₁= 4 / 1000 = 4* 10^-3 L

Also given that all of the water is converted into steam. So mass of steam that produced must have the equal mass that of water.

so mass of the steam is = 4 k.g

we know density of steam is = 0.6 k.g/m3

so volume of the steam ptoduced is = V₂ = 4 /0.6 = 6.66 liters

Here given pressure is = P = 1 atm = 101325 pascal ( s.i unit)

Now work done is defined as W = P(V₂ -V₁)

So W = 101325 (6.66 - 0.004 ) = 6683397 joule = 668.3397 kilo joule (ANSWER)

B.

We know latent heat of vapourisation of water is = 2260 kj/kg

Here we have 4 k.g of water. So amount of heat required to to convert '4' kg of water into steam is

Q = 4 * 2260 =9040 kj

Now change in internal energy is defined as

$\Delta$ E = Q -W

   $\Delta$ E =9040 - 668.3397 = 8371.6603 kilo jooule ( ANSWER )