An object with a mass of #120 g# is dropped into #640 mL# of water at #0^@C#. If the object cools by #48 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jan 29, 2018

Answer:

0.33 cal/gºC

Explanation:

At equilibrium, the final temperature #T# of both the object and the mass of water must be the same. In addition, in this condition:

#sumQ = 0#,

where #Q# indicates the amount of heat that is absorbed/lost by the two bodies in the system.

Since #Q = mcDeltaT#, then:

#Q_(object) + Q_(water) = 0#;

#m_(object)c_(object)DeltaT_(object) + m_(water)c_(water)DeltaT_(water) = 0#.

Then:

#120 g * c_(object) * (-48^oC) + 640 g * 1 (cal)/(g^oC) * 3^oC = 0#,

where I have used the density of water (1 g/ml) to estimate its mass, and #c_(water) = 1 (cal)/(g^oC).#

Finding the value for #c_(object):#

#c_(object) = (1920)/(5760)(cal)/(g^oC)#;

#c_(object) = 0.33 (cal)/(g^oC)#, approximately.