An object with a mass of 110 g is dropped into 750 mL of water at 0^@C. If the object cools by 70 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Mar 27, 2017

C_"object"=1.22"J/g°C"

Explanation:

This question is based on the concept that "heat lost"="heat gained". That is, all the heat lost by the object is gained by the water (q_"water"=q_"object").

The equation to find the change in heat is q=m*C*DeltaT, and C_"water"=4.184"J/g°C".

Since no density or mass for water is given, it can be assumed that rho_"water"=1"g/mL" so there are 750"g" of water.

Using the change in heat equation, q_"water"=m_"water"*C_"water"*DeltaT_"water"
q_"object"=m_"object"*C_"object"*DeltaT_"object"

Since q_"water"=q_"object", these two equations can be set equal to each other.
m_"water"*C_"water"*DeltaT_"water"=m_"object"*C_"object"*DeltaT_"object"

Solving this for the desired variable,
C_"object"=(m_"water"*C_"water"*DeltaT_"water")/(m_"object"*DeltaT_"object")

And plugging in the numbers
C_"object"=((750"g")(4.184"J/g°C")(3"°C"))/((110"g")(70"°C"))

Using a calculator
C_"object"=1.22"J/g°C"