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zmey [24]
3 years ago
7

If 1.9 kJ of heat is transferred to 96 g aluminum at 113°C, what would the

Chemistry
1 answer:
erastova [34]3 years ago
4 0

Answer:

T2 = 135.1°C

Explanation:

Given data:

Mass of water = 96 g

Initial temperature = 113°C

Final temperature = ?

Amount of energy transfer = 1.9 Kj (1.9×1000 = 1900 j)

Specific heat capacity of aluminium = 0.897 j/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Now we will put the values in formula.

Q = m.c. ΔT

1900 j = 96 g × 0.897 j/g.°C × T2 - 113°C

1900 j = 86.112 j/°C × T2 - 113°C

1900 j / 86.112 j/°C = T2 - 113°C

22.1°C + 113°C =  T2

T2 = 135.1°C

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Answer:

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The conductance of the electric current through the electrolytic solution increases with
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<h3>What is electrolytic solution?</h3>

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The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3
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Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Where,

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol  

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The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (280 + 273.15) K = 553.15 K  

T_2=553.15\ K  

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The conversion of T( °C) to T(K) is shown below:

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\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)

E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol

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<u>The activation energy for this reaction = 23 kJ/mol.</u>

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