Larry Finkelstein, Norman Fischer, and Cassius Schwartz have been overlooked, in my opinion.
Answer:
The average rate of energy transfer to the cooker is 1.80 kW.
Explanation:
Given that,
Pressure of boiled water = 300 kPa
Mass of water = 3 kg
Time = 30 min
Dryness friction of water = 0.5
Suppose, what is the average rate of energy transfer to the cooker?
We know that,
The specific enthalpy of evaporate at 300 kPa pressure
We need to calculate the enthalpy of water at initial state
We need to calculate the enthalpy of water at final state
Using formula of enthalpy
Put the value into the formula
We need to calculate the rate of energy transfer to the cooker
Using formula of rate of energy
Put the value into the formula
Hence, The average rate of energy transfer to the cooker is 1.80 kW.
Answer
given,
change in enthalpy = 51 kJ/mole
change in activation energy = 109 kJ/mole
when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.
where as activation energy of the product and the reactant decreases.
example:
ΔH = 51 kJ/mole
E_a= 83 kJ/mole
here activation energy decrease whereas change in enthalpy remains same.
Answer:
A
Explanation:
voltage between A and C is equal battery's voltage.
Answer:
Explanation:
Electric field due to charge at origin
= k Q / r²
k is a constant , Q is charge and r is distance
= 9 x 10⁹ x 5 x 10⁻⁶ / .5²
= 180 x 10³ N /C
In vector form
E₁ = 180 x 10³ j
Electric field due to q₂ charge
= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²
= 30.33 x 10³ N / C
It will have negative slope θ with x axis
Tan θ = .5 / √.5² + .8²
= .5 / .94
θ = 28°
E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j
= 26.78 x 10³ i - 14.24 x 10³ j
Total electric field
E = E₁ + E₂
= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j
= 26.78 x 10³ i + 165.76 X 10³ j
magnitude
= √(26.78² + 165.76² ) x 10³ N /C
= 167.8 x 10³ N / C .
