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Dovator [93]
3 years ago
9

Espresso is a coffee beverage made by forcing steam through finely ground coffee beans. Modern espresso makers generate steam at

very high pressures and temperatures, but in this problem we'll consider a low-tech espresso machine that only generates steam at 100?C and atomospheric pressure--not much good for making your favorite coffee beverage.The amount of heat Q needed to turn a mass m of room temperature ( T1) water into steam at 100?C ( T2) can be found using the specific heat c of water and the heat of vaporization Hv of water at 1 atmosphere of pressure.Suppose that a commercial espresso machine in a coffee shop turns 1.50 kg of water at 22.0?C into steam at 100?C. If c=4187J/(kg??C) and Hv=2,258kJ/kg, how much heat Q is absorbed by the water from the heating resistor inside the machine?Assume that this is a closed and isolated system.Express your answer in joules to three significant figures.Q = _________________ J
Physics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

Q = 3877 KJ

Explanation:

Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:

Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)

Q = Q₁ + Q₂   ----- equation (1)

Now, for Q₁:

Q₁ = m C ΔT

where,

m = Mass of Water = 1.5 kg

C = Specific Heat of Water = 4187 J/kg.°C

ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C

Therefore,

Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)

Q₁ = 490 x 10³ J =490 KJ

Now, for Q₂:

Q₂ = m H

where,

m = Mass of Water = 1.5 kg

H = Heat of Vaporization of Water = 2258 KJ/kg

Therefore,

Q₂ = (1.5 kg)(2258 KJ/kg)

Q₂ = 3387 KJ

Substituting the values in equation (1), we get:

Q = Q₁ + Q₂

Q = 490 KJ + 3387 KJ

<u>Q = 3877 KJ</u>

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dalvyx [7]

Answer:

An acceleration of 5m/s^2 means that the velocity of a body is increasing by 5m/s per second in a certain direction

Explanation:

5 0
3 years ago
A 1 kg box is sitting on a flat horizontal surface that has a coefficient of static friction value of 0.8. What answer best stat
Alja [10]

Answer:

c

Explanation:

5 0
3 years ago
A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?
Ivanshal [37]
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration g=9.81 m/s^2. Calling h its height at t=0, the height at time t is given by
h(t)=h- \frac{1}{2}gt^2
We are told thatn when t=0.75 s the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
0=h- \frac{1}{2}gt^2
h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m

2) The speed of the grapefruit at time t is given by
v(t)=v_0 +gt
where v_0=0 is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s
3 0
4 years ago
Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,
REY [17]

Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0
4 years ago
4000 J of heat energy is applied to a 70 g sample of water initially at 35oC. What is the final temperature of the sample?
BARSIC [14]

Answer:

48.7°C

Explanation:

Step one:

given data

Quantity of heat = 4000J

mass of sample= 70g= 0.07kg

initial temperature T1=35°C

Water has a specific heat capacity of 4182 J/kg°C

Required

The final temperature T2

Step two:

Q= mcΔT

4000= 0.07*4182(T2-35)

4000=292.74(T2-35)

4000=292.74T2-10245.9

collect like terms

4000+10245.9=292.74T2

14245.9= 292.74T2

divide both sides by 292.74

T2= 14245.9/ 292.74

T2=48.7°C

8 0
3 years ago
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