i am going crazy trying to figure this out. i just completed an experiment with heats of fusion and vaporization but for some reason i cant answer and explain this question
would 20.0g of steam at 100 degree celsius be enough to melt 20.0g of ice at 0 degrees celsius?
please break it down so i understand!!
20 g steam x heat vap = q1
20 g ice x heat fusion = q2
Is q1 > or < q2? There's you answer.
huh? im confused. whats the heat of vap? is that q=msdeltaT?
To answer the question, we need to determine if the amount of heat energy released by the steam (during condensation) is enough to melt the ice.
To do this, we can use the following formula:
Q = m * ΔH
where:
Q - heat energy in Joules
m - mass of the substance in grams
ΔH - heat of fusion or vaporization in J/g
First, we need to calculate the heat energy (Q) released by the steam.
Given:
Mass of steam (m) = 20.0g
Heat of vaporization (ΔH) = 40.7 J/g (The heat of vaporization for water)
Q(steam) = m(steam) * ΔH(vaporization)
Q(steam) = 20.0g * 40.7 J/g
Q(steam) = 814 J
Now, let's calculate the heat energy required to melt the ice:
Given:
Mass of ice (m) = 20.0g
Heat of fusion (ΔH) = 334 J/g (The heat of fusion for water)
Q(ice) = m(ice) * ΔH(fusion)
Q(ice) = 20.0g * 334 J/g
Q(ice) = 6680 J
Now, compare the two calculated values of Q.
If Q(steam) is greater than or equal to Q(ice), then there is enough heat energy released by the steam to melt the ice.
In this case, Q(steam) < Q(ice) (814 J < 6680 J)
Therefore, 20.0g of steam at 100 degrees Celsius is not enough to melt 20.0g of ice at 0 degrees Celsius.