you put 0.025 kg of water at 21.5 degrees C into a kettle and convert it all into steam. How much energy was required to heat the water to 100.0 degrees C?

Question

I need to know how to solve this problem. I have an exam tomorrow and i need to know how to do this type of problem. 

Answer 1

q = mC∆T

Step 1:  heat required to raise temp of 0.025 kg water from 21.5 degrees to 100 degrees

q = (25 g)(4.184 J/g deg)(78.5 degrees) = 8211 J

Assuming that you need to find the energy for complete conversion to steam, you would proceed as follows:

Step 2:  heat required to convert 0.025 kg water at 100 degrees to steam at 100 degrees

q = (m)(∆Hvap) = (25 g)(2260 J/g) = 56,500 J

Step 3:  add the heat from both steps:

8211 J + 56,500 J = 64,711 J = 64.7 kJ

Answer 2

Q = amount of heat that is required

 

Q = mcΔT

 

m = mass of the water = 0.025 kg

c = specific heat of water [look it up in your physics or chemistry book, in J/(kg°C)]

ΔT = temperature rise of the water = (100.0 - 21.5)°C

 

Plug in the numbers and get Q.  Note this is only the heat to raise the temperature to 100°C, as asked for in the question.  Additional heat is required to turn the 100°C water into steam, called the latent heat of vaporization, but that is a separate calculation.