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

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

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Answer 1 · 2016-10-24T15:14:45

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

$T_{f} = \frac{m_{w} C_{w} T_{i w} + c_{m} T_{i m}}{c_{m} + m_{w} C_{w}}$ $T_(f) = [m_wC_wT_(iw) + c_mT_(im)]/(c_m + m_wC_w)$

I get ${30.81}^{\circ} C$ $30.81^@ "C"$ , using $C_{w} = {4.184 J/g}^{\circ} C$ $C_w = "4.184 J/g"^@ "C"$ and $c_{m} = {55 J/}^{\circ} C$ $c_m = "55 J/"^@ "C"$ for the metal. Here we have $T_{i w} = 25^{\circ} C$ $T_(iw) = 25^@ "C"$ and $T_{i m} = 75^{\circ} C$ $T_(im) = 75^@ "C"$ .

You know that there is conservation of energy, so

$q_{m} + q_{w} = 0$ $q_m + q_w = 0$ ,

$q_{m} = - q_{w}$ $q_m = -q_w$

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

$q_m = m_mC_mDeltaT_m$
$q_w = m_wC_wDeltaT_w$

where $m$ is mass in $"g"$ , $C$ is specific heat capacity in $"J/g"^@ "C"$ or $"J/g"cdot"K"$ , and $DeltaT$ is the change in temperature in either $""^@"C"$ or $"K"$ .

$m_mC_mDeltaT_m = -m_wC_wDeltaT_w$

The metal is hotter, so dropping it into water must cool it down. So, $T_f > T_(im)$ , while $T_f < T_(iw)$ . That assures you a positive answer in the end:

$m_mC_m(T_f - T_(im)) = -m_wC_w(T_f - T_(iw))$

$m_mC_m(T_f - T_(im)) = m_wC_w(T_(iw) - T_f)$

Distribute the terms:

$m_mC_mT_f - m_mC_mT_(im) = m_wC_wT_(iw) - m_wC_wT_f$

Now get the $T_f$ terms to one side and the $T_i$ terms to the other side.

$m_mC_mT_f + m_wC_wT_f = m_wC_wT_(iw) + m_mC_mT_(im)$

Finally, factor out $T_f$ and divide:

$T_(f)(m_mC_m + m_wC_w) = m_wC_wT_(iw) + m_mC_mT_(im)$

$T_(f) = [m_wC_wT_(iw) + m_mC_mT_(im)]/(m_mC_m + m_wC_w)$

$color(blue)(T_(f) = [m_wC_wT_(iw) + c_mT_(im)]/(c_m + m_wC_w))$

where $m_mC_m = c_m$ , the heat capacity in $"J/"^@ "C"$ .

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

Answer 2 · 2016-10-24T15:27:46

Let the final temperature be $t^@C$

Heat lost by metal $=55"J/"^@Cxx(75-t)^@C$

$=55(75-t)J$

Heat gained by water
$=185gxx4.2J/(g^@C)xx(t-25)^@C$

$color(blue)("Where "4.2J/(g^@C)" is the sp.heat of water")$

So by calorimetric principle

$185xx4.2xx(t-25)=55xx(75-t)$

Now dividing both sides by 5 we get

$=>37xx4.2xx(t-25)=11(75-t)$

$=>37xx4.2t+11t=11xx75+37xx4.2xx25$

$=>166.4t=4710$

$=>t=4710/166.4~~28.3^@C$

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

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