Question

4.7 cm3 of water is boiled at atmospheric pressure to become 4063.2 cm3 of steam, also at atmospheric pressure.

1. Calculate the work done by the gas during this process. The latent heat of vaporization of water is 2.26 × 106 J/kg . Answer in units of J.

2. Find the amount of heat added to the water to accomplish this process. Answer in units of J.

3. Find the change in internal energy. Answer in units of J.

Answer 1

1.

Given,

Volume of water, V_w = 4.7 cm³ = 4.7 * 10^-6 m³

Volume of steam, V_s = 4063.2 cm³ = 4063.2 * 10^-6 m³

P_w = P_s = 1 [atm]

= 101.325 [kPa] = 101,325 [Pa]

Latent heat of vaporization, H_w = 2.26 * 10⁶ J/kg

The work done, W = PΔV

W = P (V_s - V_w)

=  101,325 Pa * (4063.2 * 10^-6 - 4.7 * 10^-6)

= 411.22 J

2)

The amount of heat added, Q = m * H_w

mass, m = density * volume

The density of water = 1000 kg/m³

=> m = 1000 * 4.7 * 10^-6

= 0.0047 kg

Q = m * H_w

  = 0.0047 * 2.26 * 10⁶

= 10622 J

3. The change in internal energy,

ΔU = Q - W

= 10622 - 411.22

= 10210.78 J

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answered by: anonymous