[eleventhxhour]

So I just started learning about enthalpy changes and all the equations that go along with it (for grade 12 chem). I wasn't sure how to do a few questions, so I'd really appreciate the help. I'm mostly kinda confused on when to use the equations q = mcΔH and ΔH=nΔHx

Here are the questions I had trouble with:

1) How much will the temperature of a 180.0g cup of coffee at 95.0°C be reduced when a 45.0g silver spoon at 25.0°C is placed in the coffee? The specific heat capacity of silver is 0.24J/(g°C)?

And I also had this question, if you have time to help with it as well:

2) The body generates energy from food and fats by the same overall process as combustion. Determine the energy per gram of glucose, C6H12O6.

C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(l) ΔH = -2803kJ

[happysmiles364]

Have you attempted the question? You need to post a attempt at it first before we can help I think.

[eleventhxhour]

Well that's the problem. I'm not sure how to start either of them. For the first one, I thought that you'd use the equation q=mcdeltaT, but you don't have the c of the cup of coffee so that doesn't make sense...

And for the second one I'm not really sure. It's probably something with converting it into moles and using the equation's mole ratios?

[Corribus]

For the first one, I think it's safe to assume that the specific heat capacity of coffee is similar to that of pure water - which you should know - assuming coffee is a sufficiently dilute solution.

For the second, if the combustion enthalpy is -2803 kJ per mole of glucose, how do you convert it to per gram?

[eleventhxhour]

For the first one, I think it's safe to assume that the specific heat capacity of coffee is similar to that of pure water - which you should know - assuming coffee is a sufficiently dilute solution.

For the second, if the combustion enthalpy is -2803 kJ per mole of glucose, how do you convert it to per gram?

Thanks! I did that and got the correct answers. (: