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valina [46]
3 years ago
12

The specific heat capacity of a certain type of cooking oil is 1.75 J/(g⋅∘C). What is the amount of heat exchanged when the temp

erature of 2.76 kg of this oil is cooled from 191∘C to 23 ∘C?
Chemistry
1 answer:
anygoal [31]3 years ago
6 0

Answer:

Q = -811440 J

Explanation:

Given data:

Mass of oil = 2.76 Kg (2.76× 1000 = 2760 g)

Initial temperature = 191 °C

Final temperature = 23°C

Specific heat capacity of oil = 1.75 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 = 23°C - 191 °C

ΔT = -168°C

Q = 2760 g ×1.75 J/g.°C  ×-168°C

Q = -811440 J

Negative sign show heat is released.

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3 years ago
For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
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Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

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