Question

If the following data is collected in Part A, what is the heat capacity of the calorimeter?

Ti cool water	21.7
Ti warm water	57.9
Tf	36.9

Answer 1

Heat of water is absorbed by the calorimeter

If Tf =final temperature of calorimeter and water, and considered that water was warm initially

Let change in temperature, of calorimeter= ∆T1=Tf-Tcool=36.9-21.7=15.2C

Change in temp, of water=∆T2=Tf-Twarm =36.9-57.9=-21 C

1) Heat of water after reaction =Q(water)

Q(water)=-Q(calorimeter)

Q(calorimeter)=C(cal)*∆T1

C(cal)=specific heat of calorimeter

Also Q(water)=(mass of water ) *C(water)* ∆T2

-C(cal)*∆T1=(mass of water ) *C(water)* ∆T2

-C(cal)=(mass of water ) *C(water)* ∆T2/∆T1

-C(cal)=(mass of water ) *C(water)*( -21/15.2)

Or, C(cal)=(mass of water ) *4.18 J/g C* 1.38=(mass of water ) *5.775 J/g C

Or, C(cal)= =(mass of water ) *5.775 J/g C