A metal sample with a mass of 50.0 grams is placed in a pot of boiling water unt

Question

A metal sample with a mass of 50.0 grams is placed in a pot of boiling water until they reach an equilibrium temperature of 100 degrees Celsius. The sample is then transferred to an insulating container full of 25.0 grams of water at a temperature of 22 degrees Celsius. They reach an equilibrium temperature of 34.59 degrees Celsius. (a) What is the temperature change of the metal sample? (b) What is the temperature change of water in the insulating cup? (c) How much heat is gained by the 25.0 grams of water? (d) How much heat is lost by the metal sample? (e) Use the information above to solve for the specific heat of the metal sample.

Explanation / Answer

-Qmetal = Qwater

-mmetal * Cpmetal * (Tf-Tmetal) = mwater*Cpwater*(Tf-Twater)

substitute

-50* Cpmetal * (34.59 -100) = 25*4.184*(34.59 -22)

solve for Cpmetal

Cpmetal = 25*4.184*(34.59 -22)/ (34.59 -100) /-50

Cpmetal = 0.402664 J/gC

a) dT for metal

Tf-Tmetal = 34.59-100 = -65.41°C

b)

dT water = (34.59-22) = 12.59 °C

c)

Qwater = 25*4.184*(34.59 -22)

Qwater= 1316.914 J

d)

Qlost = -1316.914

e)

Cp from preivous data

Cpmetal = 0.402664 J/gC

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