Chemistry, 28.10.2020 05:30 joc223

An insulated container is used to hold 47.0 g of water at 23.5°C. A sample of copper weighing 10.3 g is placed in a dry test tube and heated for 30 minutes in a boiling water bath at 100.0°C. The heated test tube is carefully removed from the water bath with laboratory
tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is
0.385 J/(g.°C), calculate the maximum temperature of the water in the insulated container after the copper metal is added.

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Answer from: GoAheadAndSmileToday

$T_f=25.0\°C$

Explanation:

Hello.

In this case, considering that the sample of hot copper is submerged into the water and the container is isolated, the heat lost by the copper is gained by the water so we can write:

$Q_{Cu}=-Q_w$

In terms of mass, specific heat and temperature we write:

$m_{Cu}C_{Cu}(T_f-T_{Cu})=-m_wC_w(T_f-T_w)$

Whereas the final temperature is the same for both copper and water because they are in contact until thermal equilibrium is reached. In such a way, the required maximum temperature no more than the equilibrium temperature and is computed as shown below:

$T_f=\frac{m_{Cu}C_{Cu}T_{Cu}+m_wC_wT_w}{m_{Cu}C_{Cu}+m_wC_w}$

Thus, plugging the given data in the formula, we obtain:

$T_f=\frac{10.3g*0.385\frac{J}{g\°C}*100\°C +47.0g*4.184\frac{J}{g\°C}*23.5\°C }{10.3g*0.385\frac{J}{g\°C}+47.0g*4.184\frac{J}{g\°C}}\\\\T_f=25.0\°C$

Which is a small change considering the initial one, because the mass of water is greater than the mass of copper as well as for the specific heats.

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Answer from: Quest

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Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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An insulated container is used to hold 47.0 g of water at 23.5°C. A sample of copper weighing 10.3 g...