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2 years ago

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How much heat does a 100. g sample of copper absorb when its temperature increases by 30.0°C? The specific heat of copper is 0.3

9 J/g °C.
a) 2340 J
b) 1170 J
c) 2.34 J
d) 1.17 J

1 answer:

erica [24]2 years ago

7 0

Answer:

$\boxed {\boxed {\sf B. \ 1170 \ Joules }}$

Explanation:

We are asked to find how much heat a sample of copper absorbs when the temperature is increased.

Since we know the mass, temperature increase, and specific heat capacity, we can use the following formula to calculate heat.

$q= mc \Delta T$

The mass of the copper sample is 100 grams, the temperature is changed or increased by 30.0 degrees Celsius, and the specific heat of copper is 0.39 Joules per gram degrees Celsius.

m= 100 g
c= 0.39 J/g °C
ΔT= 30.0 °C

Substitute the values into the formula.

$q= (100 \ g )(0.39 \ J/g \textdegree C ) (30.0 \textdegree C )$

Multiply the first two values. Note that the units of grams cancel.

$q= 39 \ J/ \textdegree C (30.0 \textdegree C )$

Multiply again, this time the units of degrees Celsius cancel.

$q= 1170 \ J$

The copper sample absorbs <u>1170 Joules</u> of heat and <u>Choice B </u>is correct.

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$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

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$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

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However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

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$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

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Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

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