Question

35.5 grams of an unknown substance is heated to 103.0 degrees Celsius and then placed into a calorimeter containing 100.0 grams of water at 24.0 degrees Celsius. If the final temperature reached in the calorimeter is 29.5 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).

Can ANYONE help with this, I am sooo lost :(

Answer 1

q lost= q gained
The Specific Heat formula is:
c = ΔQ / (m × ΔT)
q=cmΔT
Where:
c: Specific Heat Capacity
ΔQ: Heat required for the temperature change, in J
ΔT: Temperature change, in K
m: Mass of the object, in g

Answer 2

I agree with @taramgrant

q lost=q gained
The specific Heat formula
\[\Delta{Q}=mc \Delta{T}\]
where c= Specific Heat Capacity
\[\Delta{Q}=\] Heat required for the temperature change or Thermal energy change
\[\Delta{T}=\] Change in Temperature
m= Mass of the object

In this question you have to convert Celsius to Kelvin, 1 Celsius=273 kelvin
here

35.5*c*(ΔT of Unknown Substance)=100*4.18*(ΔT of Water)

Answer 3

@arindameducationusc the temperature doesn't actually have to be in kelvin because it's a change in temperature it'll be the same in both kelvin and celcius but the temperature has to be in celcius because the specific heat capacity units stated in the question are J/(°C x g) so it has to be in celcius to cancel out the unit

Answer 4

I realized I also put kelvin now too but I was copying the post from the question I answered in the physics section yesterday

Answer 5

Yes @taramgrant, because Temperature relation is very similar between kelvin and celsius.
So, in order to suppose solve this question in kelvin, Is it possible to change the Specific heat of water to another unit? in this case about 4200

Answer 6

No problem mistakes do happen sometimes..... ;)

Answer 7

To be exact 4180.

Answer 8

Well you can switch it to kelvin I guess it's just extra work lol I know in some cases like the one that I had done in the physics section had units of kelvin and it would be the same sort of solving just in kelvin

Answer 9

I got it......

Answer 10

Nice work......... @taramgrant0543664

Answer 11

Thanks