Answer:
100,200J of heat is required to convert 0.3kg of ice of 0°C to water at same temperature.
Explanation:
Heat = mass * lf
Latent heat of fusion (lf) of water is 334J/g
Heat = 300g * 334 J/g
Heat = 100,200J of heat
Gravitational acceleration is approx 9.8 m/s
Time is 7s
a=9.8 m/s
t=7s
a = d/t^2
therefore:
d = a * t^2
d = 9.8 * 7^2
d = 9.8 * 49
d = 480.2 [m]
He used truffula trees to make thneeds
Answer:
Distance = 6.667 kilometres
Explanation:
Given the following data;
Speed = 20 km/h
Departure time = 7:00
Arrival time = 7:20
Time taken = 20 minutes
To calculate the distance travelled from home to school;
First of all, we would have to convert the value of time in minutes to hours.
Conversion:
60 minutes = 1 hour
20 minutes = X hours
Cross-multiplying, we have;
X = 20/60 = 1/3 hours
Mathematically, the distance travelled by an object is calculated by using the formula;
Distance = speed * time
Distance = 20 * 1/3
Distance = 20/3 =
Distance = 6.667 kilometres
Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus, 
= = 171.82 kJ/kg
We know COP of heat pump
COP = 
= 
= 6.076
Therefore, Work out put, W = 
= 171.82 / 6.076
= 28.27 kJ/kg
