Answer:
The work done by this engine is 800 cal
Explanation:
Given:
100 g of water
120°C final temperature
22°C initial temperature
30°C is the temperature of condensed steam
Cw = specific heat of water = 1 cal/g °C
Cg = specific heat of steam = 0.48 cal/g °C
Lw = latent heat of vaporization = 540 cal/g
Question: How much work can be done using this engine, W = ?
First, you need to calculate the heat that it is necessary to change water to steam:
Here, mw is the mass of water
Now, you need to calculate the heat released by the steam:
The work done by this engine is the difference between both heats:
Answer:
1.73 seconds
Explanation:
The velocity the ball first hits the ground with is:
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (-10 m/s²) (-20 m)
v = -20 m/s
The velocity it rebounds with is 3/4 of that in the opposite direction, or 15 m/s.
The time it takes to return to the ground is:
Δx = v₀ t + ½ at²
0 = (15 m/s) t + ½ (-10 m/s²) t²
0 = t (15 − 5t²)
t = √3
t ≈ 1.73 seconds
North: 1m
South: 0,8m
Direction:
1m>0,8m
so mouse moved north
Distance:
1m-0,8m=0,2m
so mouse traveled 0,2m
Answer: The mouse moved 0,2m north.
"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."
Regards M.Y.
4.65 × 10⁴ Joules of heat is released upon converting one mole of steam to water.
<h3>Further explanation</h3>
Specific Heat Capacity is the amount of energy needed to raise temperature of 1 kg body for 1°C.
<em>Q = Energy ( Joule )</em>
<em>m = Mass ( kg ) </em>
<em>c = Specific Heat Capacity ( J / kg°C ) </em>
<em>Δt = Change In Temperature ( °C )</em>
Let us now tackle the problem!
<u>Given:</u>
initial temperature of steam = t = 100.0°C
specific heat capacity of water = c = 4.186 J/gK
mass of steam = m = 18.0 gram
final temperature of water = t' = 25.0°C
specific latent heat of vaporization of water = Lv = 2268 J/g
<u>Asked:</u>
heat released = Q = ?
<u>Solution:</u>
![\boxed {\large {\texttt{steam 100}^oC \overset{[Q_1]}{\rightarrow} \texttt{water 100}^oC \overset{[Q_2]}{\rightarrow} \texttt{water 25}^oC}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Clarge%20%7B%5Ctexttt%7Bsteam%20100%7D%5EoC%20%5Coverset%7B%5BQ_1%5D%7D%7B%5Crightarrow%7D%20%5Ctexttt%7Bwater%20100%7D%5EoC%20%5Coverset%7B%5BQ_2%5D%7D%7B%5Crightarrow%7D%20%5Ctexttt%7Bwater%2025%7D%5EoC%7D%7D)
<h3>Learn more</h3>
<h3>Answer details </h3>
Grade: College
Subject: Physics
Chapter: Thermal Physics
Keywords: Heat , Temperature , Block , Aluminium , Ice , Cold , Water
Answer: 0.42watts
Explanation:
Energy of the battery = 500 J
Time = 1200 seconds
Power of the instrument = ?
Recall that power is the rate of energy expended in doing work. Thus, power is energy expended divided by time taken.
i.e Power = (energy/time)
Power = 500J/1200 seconds
Power = 0.416 watts (Round to the nearest hundredth which is 0.42 watts)
Thus, the power of the instrument is 0.42watts
