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 × g).

HELP .

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 × g).

HELP .

aaronq · Answer

you need to use $q=m*C_p*\Delta T$, and because the heat lost by the unknown substance was transferred to the water:  $-q_{unknown}=q_{water}$

so   $m_{water}*C^{water}_p*(T_2-T_1)=q_{water}=- q_{unknown}=-[m_{unknown }*C^{unknown}_p*(T_2-T_1)]$

which is:     $m_{water}*C^{water}_p*(T_2-T_1)=-[m_{unknown }*C^{unknown}_p*(T_2-T_1)]$

they tell you everything in the question except $C^{unknown}_p$, so solve for that.

sette · Answer

I don't understand the formula u have given me. Can u talk it through for me? like explain?

aaronq · Answer

you've never seen the calorimetry formula $q=m*C_p*\Delta T$ ?

aaronq · Answer

m=mass, $C_p$= specific heat capacity, $\Delta T=T_f-T_i$ = change in temperature