Question

500g sample of water was heated with a total of 544620j of energy at 20 degree celsius. It is heated to boiling. Determine the mass of water that boiled away

Answer 1

167 g

Step-by-step explanation:

We would firstly need to understand the phase changes happening during this process:

• the sample of water is heated until its boiling point firstly: we'll calculate the amount of heat required to reach the boiling point. This will be done using the specific heat capacity of water by
  `Q = cm\Delta T`;
• the remaining amount of heat will be used to evaporate some mass of water.

Firstly, the normal boiling point of water is
`100^oC`. Find the amount of heat required to heat the sample of water:

`Q = 4.184 (J)/(g^oC)\cdot 500 g\cdot (100^oC - 20^oC) = 167360 J`

Now, find the amount of heat remaining:

`Q_(left) = 544620 J - 167360 J = 377260 J`

Let's use the equation for heat needed to evaporate water:

`Q = \Delta H^o_(vap) m`

Here the enthalpy of vaporization is:

`\Delta H^o_(vap) = 2259 J/g`

Use the amount of heat left to solve for the mass evaporated:

`m = (Q_(left))/(\Delta H^o_(vap)) = (377260 J)/(2259 J/g) = 167 g`