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Answer:
Explanation:
To solve this problem we have to take into account that the energy consumed per second is the power. Hence, by multipling the power and the time spent to arrive to the lab we obtain the total energy consumed.
But first we have to calculate the time
Now we use E=W*t for both times
A. Hence, by running the energy consumed is lower.
B.
E1=1008000J
E2=1392000J
C. Because the more intense exercise is made in a lower time in comparison with the less intense exercise, and higher the time, more energy is consumed.
Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, m₁ = 581 g
The second (mass) balance reading (of the water) in grams, m₂ = 526 g
The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ
The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ΔQ/Δm
∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g
Answer: the answer is D ☺
Explanation: