Question

31.5 grams of an unknown substance is heated to 102.4 degrees Celsius and then placed into a calorimeter containing 103.5 grams of water at 24.5 degrees Celsius. If the final temperature reached in the calorimeter is 32.5 degrees Celsius, what is the specific heat of the unknown substance?
Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(°C x g)

Answer 1

By conservation of energy:
\[Heat\:gained\:by\:water=Heat\:lost\:by\:unknown\:substance\]\[4.18*103.5*(32.5-24.5)=c(31.5)(102.4-32.5)\]

Solve for c to get the specific heat capacity of the unknown substance in J/(°C g)

If you don't know where the equation came from just ask.

Answer 2

would you please explain

Answer 3

heat gained or lost in a substance is simply: mass x specific heat capacity x change in temperature

This is because as we have more stuff to heat up, we need more heat. But different substances heat up differently, so there is a specific heat capacity to distinguish them and describe how easy it is to heat them or cool them.

Answer 4

thanks. you're exetremely helpful

Answer 5

what is c suppose to equal?

Answer 6

@Festinger

Answer 7

1.57J/(g K)

Answer 8

i didn't mean the answer lol. i meant like it stands for

Answer 9

read my first answer again:
c = specific heat capacity of the unknown substance in J/(°C g)
it is the measure of how much heat you need to give to the substance to bring 1g of it up by 1 degrees celcius.

Answer 10

oh ok cool

Answer 11

4.18* 103.5* (32.5-27.5) = c(31.5)(102.4-32.5)
432.63(5)=c(31.5)(69.9)
2163.15=c2201.85
c=.982
that's what i got...

Answer 12

@Festinger

Answer 13

What is the initial temperature of water? is it 24.5C or 27.5C?

Answer 14

24.5. don't you need to convert it to kelvin though

Answer 15

But your first line says: 4.18* 103.5* (32.5-27.5) = c(31.5)(102.4-32.5)

Which says 27.5

Answer 16

that's where i made my mistake