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zloy xaker [14]

3 years ago

5

Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

heat released during condensation goes only toward heating the metal, what is the final temperature (in degrees Celsius) of the metal block

1 answer:

GarryVolchara [31]3 years ago

6 0

Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

$T_{final}=25^oC$

Thus, the final temperature of the metal block is, $25^oC$

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Read 2 more answers

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

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Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

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