# What is the final temperature if a metal at 75^@ "C" is dropped into "185 g" of water starting at 25^@ "C"? The heat capacity of the metal is "55 J/g"*""^@ "C".

Oct 24, 2016

This is one of those questions where solving it in general would help you more with not messing up on the math.

${T}_{f} = \frac{{m}_{w} {C}_{w} {T}_{i w} + {c}_{m} {T}_{i m}}{{c}_{m} + {m}_{w} {C}_{w}}$

I get ${30.81}^{\circ} \text{C}$, using ${C}_{w} = \text{4.184 J/g"^@ "C}$ and ${c}_{m} = \text{55 J/"^@ "C}$ for the metal. Here we have ${T}_{i w} = {25}^{\circ} \text{C}$ and ${T}_{i m} = {75}^{\circ} \text{C}$.

You know that there is conservation of energy, so

${q}_{m} + {q}_{w} = 0$,

${q}_{m} = - {q}_{w}$

where $m$ is metal and $w$ is water, while $q$ is heat flow in $\text{J}$. That means you can equate their equations:

${q}_{m} = {m}_{m} {C}_{m} \Delta {T}_{m}$
${q}_{w} = {m}_{w} {C}_{w} \Delta {T}_{w}$

where $m$ is mass in $\text{g}$, $C$ is specific heat capacity in $\text{J/g"^@ "C}$ or $\text{J/g"cdot"K}$, and $\Delta T$ is the change in temperature in either $\text{^@"C}$ or $\text{K}$.

${m}_{m} {C}_{m} \Delta {T}_{m} = - {m}_{w} {C}_{w} \Delta {T}_{w}$

The metal is hotter, so dropping it into water must cool it down. So, ${T}_{f} > {T}_{i m}$, while ${T}_{f} < {T}_{i w}$. That assures you a positive answer in the end:

${m}_{m} {C}_{m} \left({T}_{f} - {T}_{i m}\right) = - {m}_{w} {C}_{w} \left({T}_{f} - {T}_{i w}\right)$

${m}_{m} {C}_{m} \left({T}_{f} - {T}_{i m}\right) = {m}_{w} {C}_{w} \left({T}_{i w} - {T}_{f}\right)$

Distribute the terms:

${m}_{m} {C}_{m} {T}_{f} - {m}_{m} {C}_{m} {T}_{i m} = {m}_{w} {C}_{w} {T}_{i w} - {m}_{w} {C}_{w} {T}_{f}$

Now get the ${T}_{f}$ terms to one side and the ${T}_{i}$ terms to the other side.

${m}_{m} {C}_{m} {T}_{f} + {m}_{w} {C}_{w} {T}_{f} = {m}_{w} {C}_{w} {T}_{i w} + {m}_{m} {C}_{m} {T}_{i m}$

Finally, factor out ${T}_{f}$ and divide:

${T}_{f} \left({m}_{m} {C}_{m} + {m}_{w} {C}_{w}\right) = {m}_{w} {C}_{w} {T}_{i w} + {m}_{m} {C}_{m} {T}_{i m}$

${T}_{f} = \frac{{m}_{w} {C}_{w} {T}_{i w} + {m}_{m} {C}_{m} {T}_{i m}}{{m}_{m} {C}_{m} + {m}_{w} {C}_{w}}$

$\textcolor{b l u e}{{T}_{f} = \frac{{m}_{w} {C}_{w} {T}_{i w} + {c}_{m} {T}_{i m}}{{c}_{m} + {m}_{w} {C}_{w}}}$

where ${m}_{m} {C}_{m} = {c}_{m}$, the heat capacity in $\text{J/"^@ "C}$.

So now, we can plug the numbers in and solve for the final temperature.

Oct 24, 2016

Let the final temperature be ${t}^{\circ} C$

Heat lost by metal $= 55 {\text{J/}}^{\circ} C \times {\left(75 - t\right)}^{\circ} C$

$= 55 \left(75 - t\right) J$

Heat gained by water
$= 185 g \times 4.2 \frac{J}{{g}^{\circ} C} \times {\left(t - 25\right)}^{\circ} C$

$\textcolor{b l u e}{\text{Where "4.2J/(g^@C)" is the sp.heat of water}}$

So by calorimetric principle

$185 \times 4.2 \times \left(t - 25\right) = 55 \times \left(75 - t\right)$

Now dividing both sides by 5 we get

$\implies 37 \times 4.2 \times \left(t - 25\right) = 11 \left(75 - t\right)$

$\implies 37 \times 4.2 t + 11 t = 11 \times 75 + 37 \times 4.2 \times 25$

$\implies 166.4 t = 4710$

$\implies t = \frac{4710}{166.4} \approx {28.3}^{\circ} C$