A 23.0 gram sample of an alloy was heated to 100.00 degree C and dropped into a

Question

A 23.0 gram sample of an alloy was heated to 100.00 degree C and dropped into a beaker containing 92 grams of water at 25.26 degree C. The temperature of the water rose to a final temperature of 27.84 degree C. Neglecting heat losses to the room and the heat capacity of the beaker itself, what is the specific heat of the alloy? To determine the heat capacity of a calorimeter. 50.0 mL of water at 56 degree C is added to a calorimeter containing 50.0 mL of water at 23 degree C. After about five minutes of mixing, the final temperature of the solution inside the calorimeter reached 37 degree C. Calculate the heat capacity of the calorimeter (J/degree C - Joules per degree).

Explanation / Answer

for the metal

Q1 = m1*C1p*(Tf-Tm)

for wter

Q2= m2*Cp2*(Tf-Tw)

then

Q1 = -Q2

m1*Cp1*(Tf-Tm) = -m2*Cp2*(Tf-Tw)

Cp1 =  -m2*Cp2*(Tf-Tw)/((m1*(Tf-Tm))

Cp1 = (-92*4.184*(27.84-25.26))/(23*(27.84-100)) =0.59837 J/gC

Cp = 0.59837 J/gC

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