Question

 

A. 42.0000 g of an unknown is heated to 98.69℃. It is then placed in 30.0000g of water at 20.00℃, what is the final temperature if the specific heat of the metal is 0.2202 J/(gK)?

B. If 80.0 J was used to heat 10.0g of one of the following metals, which would experience the largest temperature change?  

Al (Specific Heat = 0.900 J/(gK))

Fe   (Specific Heat = 0.450 J/(gK))

Cu   (Specific Heat = 0.387 J/(gK))

Au (Specific Heat = 0.129 J/(gK))

C. 40.3660 g of an unknown is heated to 94.20℃. It is then placed in 25.0000g of water at 25.01℃, the final temperature of the water and the metal is 42.22℃. What is the specific heat of the metal?

Answer 1

A)

heat balance

-Qhot = Qcold

-m1*C1*(Tf-T1) = m2*C2*(Tf-T2)

-42*0.2202*(Tf-98.69) = 30*4.184(Tf-20)

-0.07368*Tf = + 98.69*-0.07368 = Tf-20

Tf(-0.07368 -1) = -20 + 98.69*0.07368

Tf = -12.7285 / -1.07368

Tf = 11.85°C

b)

largest temperature change is given for the

LEAST Heat capacity...

that is, the Cp which is lower, will have higher T change since it is inversely proportional

From th elist

choose Gold, since it will increase drastically

C

similarly

-Qhot = Qcold

-m1*C1*(Tf-T1) = m2*C2*(Tf-T2)

-40.366*Cp ( 42.22-94.20) = 25*4.184 ( 42.22- 25.01 )

Cp = 25*4.184 ( 42.22- 25.01 ) / (-40.366) /  ( 42.22-94.20)

Cp = 0.85794 J/gC