42.5 grams of an unknown substance is heated to 105.0 degrees Celsius and then placed into a calorimeter containing 110.0 grams of water at 24.2 degrees Celsius. If the final temperature reached in the calorimeter is 32.4 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 × g).

Question

42.5 grams of an unknown substance is heated to 105.0 degrees Celsius and then placed into a calorimeter containing 110.0 grams of water at 24.2 degrees Celsius. If the final temperature reached in the calorimeter is 32.4 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 × g).

superhelp101 · Answer

@JoannaBlackwelder ?

superhelp101 · Answer

@aaronq ?

aaronq · Answer

we know that the thermal energy (heat) given off by the unknown substance is absorbed by the water, mathematically, it's expressed: 

                           $-q_{unknown}=q_{water}$
 
            $\underbrace{m*C_p*(T_f-T_i)}_{unknown}=\underbrace{m*C_p*(T_f-T_i)}_{water}$

aaronq · Answer

Sorry, i missed a negative sign, it should be:

$-[\underbrace{m*C_p*\Delta T}_{unknown}]=\underbrace{m*C_p*\Delta T}_{water}$

superhelp101 · Answer

hmmm so this is what i did is it wrong 
110.0g x 4.18J/(8.2 degrees Celsius) = 56.073 degrees celcius

aaronq · Answer

did you divide by 8.2 or multiply?

superhelp101 · Answer

oh i divided

aaronq · Answer

well thats clearly not what the formula says.. lol 

If you multiplied you would've found the heat absorbed by the water, which is half of the problem.

superhelp101 · Answer

oh oops i will try to solve it then i show you if it is right or wrong

aaronq · Answer

okayss

superhelp101 · Answer

1. We must know how many joules of heat must to be transferred to change the temperature of a gram of the unknown substance by a degree Celsius. So first you have to know how much heat that 42.5 g chunk gave off
2.  I’ll look like 110.0 gH2O*(4.186 J/(gH2O degreeC))*(32.4 degreesC-24.2 degreesC)=3775.775 J 
3. The units cancel to J, this is joules of heat transferred to the water. Also the 32.4 - 24.2 is the change in temperature for the water. 
4. Now that we know how many joules of heat were transferred from the substance to the water we have to use that value to find how many joules were transfer per g of substance per degree C because that will give us the specific heat in J/(g degree C). 
5. -3775.775J / (42.5 g unknown*(32.4 degrees C-105.0 degrees C)) = 1.22372 J/(g unknown degreeC) 
??

aaronq · Answer

yep! you got it. good job !

superhelp101 · Answer

thank youuuuuu