When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di
1 answer:
To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.
We know that by Hooke's law
Where,
k = Spring constant
x = Displacement
Re-arrange to find k,
Perioricity in an elastic body is defined by
Where,
m = Mass
k = Spring constant
Therefore the period of the oscillations is 0.685s
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Answer:
= 85.7 ° C
Explanation:
For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state
Q₁ = m L
Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water
Q₁ = 2.00 10⁻³ 2.26 10⁶
Q1 = 4.52 10³ J
Now the heat of coffee in the cup, which does not change state is
Q coffee = M (
-
)
Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat
Qc = - Q₁
M ce ( -
) = - Q₁
The coffee dough left in the cup after evaporation is
M = 250 -2 = 248 g = 0.248 kg
-Ti = -Q1 / M
= Ti - Q1 / M
Since coffee is essentially water, let's use the specific heat of water,
= 4186 J / kg ºC
Let's calculate
= 90.0 - 4.52 103 / (0.248 4.186 103)
= 90- 4.35
= 85.65 ° C
= 85.7 ° C