# 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 5 ^@C, what is the specific heat of the material that the object is made of?

Aug 8, 2017

C_("object")~~2.038"J"//"g"^o"C"

#### Explanation:

We can find the specific heat of the material that the object is made of by using the first law of thermodynamics and calorimetry.

The first law of thermodynamics is given by the equation:

$\textcolor{c r i m s o n}{\Delta {E}_{t h} = W + Q}$

where ${E}_{t h}$ is thermal energy, $W$ is the work performed on/by a system and $Q$ is heat transferred into/out of it

We can also express the thermal energy in terms of temperature change:

$\textcolor{s k y b l u e}{\Delta {E}_{t h} = M c \Delta T}$

where $M$ is the mass of the substance, $c$ is its specific heat, and $\Delta T$ is the change in temperature experienced by the substance

We can set $W = 0$ as we are heating the water with our object (or cooling our object with the water), not performing work on it. Combining the two equations above, we get:

$\textcolor{\in \mathrm{di} g o}{Q = M c \Delta T}$

We are provided with the following information:

• $\to \text{M"=110"g}$

• $\to \text{V"=750"mL}$

• ->(T_i)_("water")=0^o"C"

• ->(T_f)_("water")=5^o"C"

• ->(DeltaT)_("object")=-70^o"C"

It is known that the specific heat of water is $4.184 \text{J"//"g"^o"C}$. Additionally, we can find the mass of the water using its known density of $1 {\text{g"//"cm}}^{3}$, as well as the fact that $1 {\text{mL"=1"cm}}^{3}$.

750cancel("cm"^3)*(1"g")/(cancel("cm"^3))

$= \textcolor{l i g h t g r e e n}{750 \text{g}}$

We have enough information to calculate $Q$ for the water now. Because the heat lost by the object must be equal to the heat gained by the water (first law, assuming insulation), we can then use the value we obtain to find the specific heat of the material.

${Q}_{\text{water")=M_("w")c_("w")DeltaT_("w}}$

$= \left(750 \text{g")(4.184"J"//"g"^o"C")(5^o"C}\right)$

$= \textcolor{g r e y}{15690 \text{J}}$

Now, for the unknown object:

${Q}_{\text{object")=M_("o")c_("o")DeltaT_("o}}$

Solving for ${c}_{o}$:

${c}_{\text{o")=Q_("o")/(M_"o"*DeltaT_"o}}$

And since ${Q}_{\text{water")=-Q_("object}}$ (the heat gained by the water is the heat lost by the object), we have:

c_"o"=(-15690"J")/(110"g"*70^o"C")

$\approx \textcolor{\in \mathrm{di} g o}{2.038 \text{J"//"g"^o"C}}$