Energy required to raise the temperature from 35°C - 45 °C= 25116 J.
specific heat, the quantity of warmth required to raise the temperature of one gram of a substance by means of one Celsius degree. The units of precise warmth are generally energy or joules consistent with gram according to Celsius diploma. for instance, the unique warmth of water is 1 calorie (or 4.186 joules) according to gram in step with Celsius degree.
solving,
Sample of liquid = 400. 0 g
temperature = 30. 0 ºc
joules of energy are required to raise the temperature of the water to 45. 0 ºc
therefore rise in temperature 45 - 30 = 15°C
Specific heat capacity = 4.186 J/g m °C
In kelvin = 273 + 15 = 288
= ∴ energy required = Q = m s ( t final - t initial)
= 400*4.186 * 15
= 25116 joule
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The turbine would stop generating electricity
Answer:
0.00011 JK.
The process does NOT violate the second law of thermodynamics
Explanation:
The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.
=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.
The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.
The change in entropy of the cold container = 1/295 = 0.00339 J/K.
Therefore, the change in the entropy of the system = - 0.00328 J/K + 0.00339 J/K = 0.00011 JK.
Note that the change in entropy of the system gives a positive value. Hence, this process does not violate the second law of thermodynamics.
The process does NOT violate the second law of thermodynamics.
Answer:
The unbalanced chemical equation: H₂O₂ → H₂O + O₂.
The balanced chemical equation: H₂O₂ → H₂O + 1/2O₂.
Explanation:
- Hydrogen peroxide is decomposed into oxygen and water, which is a slow reaction.
- It is can be catalyzed by using yeast.
The unbalanced chemical equation: H₂O₂ → H₂O + O₂.
The balanced chemical equation: H₂O₂ → H₂O + 1/2O₂.
1.0 mol of H₂O₂ is decomposed to 1.0 mol of H₂O and 0.5 mol of O₂.
