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4 months ago

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A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temp

erature rise, assuming it does not melt?

1 answer:

lorasvet [2.7K]4 months ago

0 0

Response: The increase in temperature is $0.53^0C$

Reasoning:

The amount of thermal energy needed to elevate the temperature of a given substance by one degree Celsius is referred to as the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat gained by ice = 5280 J

m = mass of ice = 2.40 kg = 2400 g (1kg=1000g)

c = heat capacity of water = $4.18J/g^0C$

Initial temperature = $T_i$

Final temperature = $T_f$

Temperature change , $\Delta T=T_f-T_i=?$

Substituting the values, we obtain:

$5280J=2400g\times 4.18J/g^0C\times \Delta T$

$\Delta T=0.53^0C$

Therefore, the temperature increase is $0.53^0C$

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• Briefly touch the balloon to a metal object. Why is this step necessary?

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PART A: Use the following glycolytic reaction to answer the question. If the concentration of DHAP is 0.125 M and the concentrat

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Answer:

For A: The change in free energy for the reaction is -5339.76 J/mol

For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

Explanation:

Regarding the specified chemical reaction:

$DHAP\rightleftharpoons G_3P$

For A:

The relationship between standard Gibbs free energy and equilibrium constant is as follows:

$\Delta G^o=-RT\ln K_{eq}$

The free energy change can be calculated using the following equation:

$\Delta G=\Delta G^o+RT\ln Q$

Or,

$\Delta G=-RT^o\ln K_{eq}+RT\ln Q$

where,

$\Delta G$ = Change in free energy

R = Gas constant = $8.314J/K mol$

$T^o$ = standard temperature = $25^oC=[273+25]K=298K$

T = temperature of the cell = $37^oC=[273+37]K=310K$

$K_[eq}$ = equilibrium constant = $5.4\times 10^{-2}$

Q = reaction quotient = $\frac{[G_3P]}{[DHAP]}$

$[G_3P]$ = 0.06 M

[DHAP] = 0.125 M

Substituting the values into the equation yields:

$\Delta G=[-(8.314J/mol.K\times 298K\times \ln (5.4\times 10^{-2}))]+[(8.314J/mol.K\times 310K\times \ln (\frac{0.06}{0.125}))]\\\\\Delta G=-[-7231.46]+[-1891.7]=-5339.76J/mol$

Thus, the change in free energy for the reaction is -5339.76 J/mol

For B:

To convert the free energy change to kilojoules, we apply the conversion factor:

1 kJ = 1000 J

So, $-5339.76J/mol\times \frac{1kJ}{1000J}=-5.34kJ/mol$

Consequently, the free energy change's units are kJ/mol

For C:

For spontaneity in the reaction, the Gibbs free energy must be negative. However, the calculations indicate a positive Gibbs free energy, leading to the conclusion that the reaction is not spontaneous.

The free energy change of the reaction is negative.

Consequently, the forward reaction is favored and the reverse reaction is not favored.

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