Consider a 10 gram sample of a liquid with specific heat 2 J/g*K. By the addition of 400 J, the liquid increases its temperature by 10 K and then boils. Half of the liquid boils away before all the heat is used up. What is the heat of vaporization of the liquid?
A. 10 J/g
B. 20 J/g
C. 80 J/g
D. 200 J/g
E. 40 J/g
- I am stuck on this!
10 grams * 2 J/gdegK = 20 J/degK
20J/degK * 10 deg K = 200 Joules to get it all to the boiling point
then boil 5 grams with remaining 200 J
heat/gram = 200 Joules/5 grams
= 40 Joules/gram
To find the heat of vaporization of the liquid, we need to first calculate the amount of heat required to raise the temperature of the liquid and then determine the remaining heat that is used to vaporize the liquid.
First, let's calculate the heat required to raise the temperature of the 10 gram sample of the liquid by 10 K using the specific heat formula:
Heat (Q) = mass (m) x specific heat (C) x change in temperature (ΔT)
Q1 = 10 g x 2 J/g*K x 10 K
Q1 = 200 J
So, 200 J of heat is used to raise the temperature of the liquid.
Now, the remaining 200 J of heat is used to vaporize half of the liquid. Since half of the liquid boils away, the mass of the liquid remaining is 5 grams (10 grams / 2).
To calculate the heat required to vaporize the remaining liquid, we use the formula:
Heat of vaporization (Q2) = mass (m) x heat of vaporization (Hv)
Q2 = 5 g x Hv
Now we can substitute the value of Q2 and the remaining heat into the equation and solve for the heat of vaporization (Hv):
200 J = 5 g x Hv
Hv = 200 J / 5 g
Hv = 40 J/g
Therefore, the heat of vaporization of the liquid is 40 J/g. So the correct answer is E. 40 J/g.