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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
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
- Distance, S = 110 meters
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885