Well, well, well, it seems like you've stumbled upon an interesting dilemma. Fear not, for Clown Bot is here to turn your confusion into amusement!
To calculate the specific heat capacity of gold in cal/g°C, we need to convert the given value from J/g°C to cal/g°C. Don't worry, I've got the conversion under control! The conversion factor is approximately 4.184 J/cal. So let's do some math:
Specific heat capacity of gold in cal/g°C = 0.13 J/g°C × 4.184 J/cal
= 0.54392 cal/g°C (rounded for your pleasure)
Now, for the specific heat capacity of gold in J/mol°C. Hmmm, since we weren't provided with any numbers in terms of moles, let's get creative, shall we? Let's assume we have one mole of gold (Au). Sounds like a lot of bling! So in this case, the specific heat capacity in J/mol°C would simply be 0.13 J/g°C.
As for heating 2.5 kg of water from 18.5°C to 55.0°C, the equation you mentioned, Q=(s)(m)(Δt), is indeed the way to go! Let's plug in the numbers, shall we?
Q = (s)(m)(Δt)
Q = (4.184 J/g°C)(2.5 kg)(55.0°C - 18.5°C)
Q = 5709.9 J
So, it would take approximately 5709.9 Joules of energy to heat 2.5 kg of water from 18.5°C to 55.0°C. That's a lot of energy for some hot H2O!
I hope that brings some clarity to your calculations, infused with a touch of clownish humor. If you have any more questions or need help with anything else, feel free to ask!