Question

26. A physical science student is performing a thermodynamics experiment in the school laboratory. She takes a 30.0 g ice cube at 0. °C and melts it to water in a beaker over a laboratory burner. She then heats the water to boiling at 100. °C. Finally she boils the water entirely to steam at 100. °C. How much thermal energy was gained by the water during this entire process?

Answer 1

The thermal energy gained by the water during the entire process is 90,372 J

Step-by-step explanation:

Given;

mass of the ice cube, m = 30 g

latent heat of fusion of ice, L = 334 J/g

specific heat capacity of water, C = 4.184 J/g⁰C

latent heat of vaporization of water, h = 2260 J/g

The thermal energy gained by the water during the entire process is calculated as;

`Q_t = Q_(fusion-ice) \ + \ Q_(boil-water) \ + \ Q_(steam-vapor)`

The heat of fusion of the ice at 0⁰C is calculated as;

`Q_(fusion) = mL = 30\ g * 334 J/g = 10020 \ J`

The heat of capacity of the water from 0⁰C to 100 ⁰C

`Q_(boil-water) = mc\Delta \theta = mc(100-0) = 30\ g * 4.184\ J/g^0C * 100\ ^0C = 12,552 \ J`

The heat of vaporization of the steam at 100⁰C is calculated as;

`Q_(steam) = mh= 30 g * 2260 \ J/g= 67,800 \ J`

The thermal energy gained by the water during the entire procees;

Qt = 10,020 J + 12,552 J + 67,800 J

= 90,372 J

Therefore, the thermal energy gained by the water during the entire process is 90,372 J