# 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?

Mar 27, 2017

${C}_{\text{object"=1.22"J/g°C}}$

#### Explanation:

This question is based on the concept that $\text{heat lost"="heat gained}$. That is, all the heat lost by the object is gained by the water (${q}_{\text{water"=q_"object}}$).

The equation to find the change in heat is $q = m \cdot C \cdot \Delta T$, and ${C}_{\text{water"=4.184"J/g°C}}$.

Since no density or mass for water is given, it can be assumed that ${\rho}_{\text{water"=1"g/mL}}$ so there are $750 \text{g}$ of water.

Using the change in heat equation, ${q}_{\text{water"=m_"water"*C_"water"*DeltaT_"water}}$
${q}_{\text{object"=m_"object"*C_"object"*DeltaT_"object}}$

Since ${q}_{\text{water"=q_"object}}$, these two equations can be set equal to each other.
${m}_{\text{water"*C_"water"*DeltaT_"water"=m_"object"*C_"object"*DeltaT_"object}}$

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
${C}_{\text{object"=1.22"J/g°C}}$