The energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
U =
m ΔT
U = Energy
= Specific heat
m = Mass
ΔT = Change in temperature
ρ = Density
V = Volume
ρ = 1000 g / m³ (Dry air )
= 1 J / g K
ΔT = 5 °C
V = 4 * 4 * 3
V = 48 m³
m = ρ V
m = 1000 * 48
m = 48000 g
U = 1 * 48000 * 7
U = 336000 J
U = 336 kJ
Therefore, the energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
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Water vapour has more kinetic energy.
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Answer:
The work done on the suitcase is, W = 1691 J
Explanation:
Given data,
The force on the suitcase is, F = 89 N
The distance Russell dragged the suitcase, S = 19 m
The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction
The work done on the suitcase by Russell is given by the formula
W = F · S
Substituting the given values,
W = 89 N x 19 m
W = 1691 J
Hence, the work done on the suitcase is, W = 1691 J