Wednesday, June 12, 2019
Unizor - Physics4Teens - Energy - Measuring Heat - Problems
Notes to a video lecture on http://www.unizor.com
Measuring Heat - Problems
How much heat energy is required to raise the temperature of 1 kg of water from 20°C to a boiling point of 100°C?
Assume the specific heat capacity of water is
Cw = 4184 J/(kg·°C).
Q = C·m·(Tend−Tbeg) =
= 4184·1·(100−20) =
= 334,720 J
A piece of unknown metal of mass Mm and temperature Tm was put into an isolated reservoir filled with Mw mass of water at temperature Tw. After the system of water and metal came to thermal equilibrium, its temperature became T.
Assume that the metal is not too hot (so, water will not vaporize) and not too cold (so, the water will not freeze).
Assuming that the water's specific heat capacity is known and equals to Cw, what is the specific heat capacity Cm of the unknown metal?
from which Cm equals to
A burger has about 300 kcal of energy in it.
1 kcal = 4184 J.
A person, who ate it, wants to spend this energy by climbing up the stairs. A person's mass is 75 kg, the height between the floor is 3 m.
Assume that only 25% of energy in the food can be used for climbing, while the other 75% is needed to maintain our body's internal functions.
Counting from the ground floor as floor #0, to what floor can a person climb using that energy from a burger?
Ebur = 300kcal · 4184J/kcal =
Eclimb = 0.25·Ebur =
Efloor = 75kg · 9.8m/sec² · 3m =
N = Eclimb / Efloor ≅ 142 floors
An ice of mass 0.1kg has temperature −10°C.
What's the minimum amount of water M at temperature 20°C needed to melt it?
Assume, specific heat capacity of water is 4183J/(kg·°C) and that of ice is 2090J/(kg·°C). Assume also that the amount needed to melt ice at 0°C is 333,000J/kg.
Ewarm = 2090·0.1·10 = 2090J
Emelt = 333000·0.1 = 33300J
Eneed = 2090 + 33300 = 35390J
Ewater = 4183·M·20 = 83660·M
Ewater = Eneed
35390 = 83660·M
M = 0.423kg