Unizor - Physics4Teens - Energy - Measuring Heat - Problems

Notes to a video lecture on http://www.unizor.com

Measuring Heat - Problems

Problem 1

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
C_w = 4184 J/(kg·°C).

Answer
Q = C·m·(T_end−T_beg) =
= 4184·1·(100−20) =
= 334,720 J

Problem 2

A piece of unknown metal of mass M_m and temperature T_m was put into an isolated reservoir filled with M_w mass of water at temperature T_w. 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 C_w, what is the specific heat capacity C_m of the unknown metal?

Answer

C_w·M_w·(T−T_w) =
= C_m·M_m·(T_m−T)
from which C_m equals to
C_w·M_w·(T−T_w)/[M_m·(T_m−T)]

Problem 3

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?

Answer
E_bur = 300kcal · 4184J/kcal =
= 1,255,200J
E_climb = 0.25·E_bur =
= 313,800J
E_floor = 75kg · 9.8m/sec² · 3m =
= 2205J
N = E_climb / E_floor ≅ 142 floors

Problem 4

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.

Answer
E_warm = 2090·0.1·10 = 2090J
E_melt = 333000·0.1 = 33300J
E_need = 2090 + 33300 = 35390J
E_water = 4183·M·20 = 83660·M
E_water = E_need
35390 = 83660·M
M = 0.423kg