How much energy in kilojoules is needed to heat 5.00 g of ice fro...

Question

How much energy in kilojoules is needed to heat 5.00 g of ice from -11.0 °C to °30.0 °C? The heat of fusion of water is 
6.01 kJ>mol, and the m# olar heat capacity is 36.01 kJ>mol for ice and 75.4 J/K mol2 for liquid water

Accepted Answer

Hello everyone today. We have the following problem, Calculate the energy and kill Eagles absorbed when .43 moles of liquid decade is heated from 132.44°C two, and completely evaporated for decades. We have a boiling point. We have the entropy of vaporization and we have our specific heat. So we know that from this temperature increase decade undergoes vaporization. So we have to calculate the total energy of this reaction or the heat that's absorbed from this heating plus the heat that is absorbed from condensation. So we have the heat that's the total. Which is this heat from absorbed from heating plus the heat from the condensation. So how do we calculate for both of these? Well, for our cue liquid That's going to be from our moles that we were given our 0.43 moles times that specific key for that. Which was 314.4 joules per mole kelvin. We then have to multiply by our degree change or a temperature change which was the final temperature or 74.15 minus the initial temperature which is 32.44 degrees Celsius. And when we do that we get a value of 5. kg jewels. Now we need to calculate from the heat that's absorbed from condensation, It's gonna be cute of our vaporization. We're gonna take the moles once again, .43 moles. We're gonna multiply by the delta h vaporization or the entropy of vaporization which is 39.58 kill it jules per mole. When our units cancel out, we're left with 17.019 killed Jules. And so we can finally calculate our total heat. So we're total heat. It's going to be our 5.639 kilo joules plus our 17.19 kg jewels. And that's gonna give us a total value of 22.7 kg jewels as our final answer. Overall, I hope this helped ahead until next time.

Watch next