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

Answer:

167 g

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:

$Q = 4.184 \frac{J}{g^oC}\cdot 500 g\cdot (100^oC - 20^oC) = 167360 J$

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:

$m = \frac{Q_{left}}{\Delta H^o_{vap}} = \frac{377260 J}{2259 J/g} = 167 g$