Chemistry, 14.08.2020 01:01 TianaTheSquid

A 12.6 g block of a metal intially at a temperature of 94.0 °C is added to a 100.0 g sample of water intially at 21.0 °C. If the final temperature of the water is 25.1 °C, determine the specific heat of the metal. Write all equations or formulas you use, and show as much of your work as you can.

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Answer from: marlesly87

The specific heat of the metal is 0.0482 joules per gram-Celsius.

Explanation:

Block of metal is cooled by adding water and thermal equilibrium is reached. According to the First Law of Thermodynamics and by supposing the absence of energy and mass interaction of the system with surroundings, the change in the energy of the system is represented by:

$\Delta U_{m} + \Delta U_{w} = 0$

Where $\Delta U_{m}$ and $\Delta U_{w}$ are the changes in internal energies of the block of metal and water, measured in joules.

The expression described above is now extended by applying the definition of internal energy for constant mass systems:

$m_{m}\cdot c_{v,m}\cdot (T_{m,in}-T_{m,out}) + m_{w}\cdot c_{v,w}\cdot (T_{w,in}-T_{w,out})= 0$

Where:

$m_{m}$ , $m_{w}$ - Masses of the block of metal and water, measured in grams.

$c_{v,m}$ , $c_{v,w}$ - Specific heats of the block of metal and water, measured in joules per gram-Celsius.

$T_{in, m}$ , $T_{in, w}$ - Initial temperatures of the block of metal and water, measured in Celsius.

$T_{out, m}$ , $T_{out, w}$ - Final temperatures of the block of metal and water, measured in Celsius.

The specific heat of the metal is cleared in the equation:

$c_{v,m} = -\frac{m_{w}\cdot c_{v,w} (T_{w,in}-T_{w,out})}{m_{m}\cdot (T_{m,in}-T_{m,out})}$

If $m_{m} = 12.6\,g$ , $m_{w} = 100\,g$ , $c_{v,w} = 4,186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,in} = 21\,^{\circ}C$ , $T_{m,in} = 94\,^{\circ}C$ and $T_{m,out} = T_{w,out} = 25.1\,^{\circ}C$ , the specific heat of the metal is:

$c_{v,m} = -\frac{(100\,g)\cdot \left(4,186\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (25\,^{\circ}C-25.1\,^{\circ}C)}{(12.6\,g)\cdot (94\,^{\circ}C-25.1\,^{\circ}C)}$

$c_{v,m} = 0.0482\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of the metal is 0.0482 joules per gram-Celsius.

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

Answer from: Quest

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A 12.6 g block of a metal intially at a temperature of 94.0 °C is added to a 100.0 g sample of water...