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Tresset [83]
3 years ago
10

How long does it take to cook potatoes in a pressure cooker?

Physics
2 answers:
Artist 52 [7]3 years ago
7 0
It can only take 15 minutes to 20 minutes

BARSIC [14]3 years ago
3 0
For sliced or pieces, as well as new or small potatoes, about 5-7 minutes. For whole medium-big potatoes, 10-12 minutes 
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1.3 A
If a clock expends 2 W of power from a 1.5 V battery, what amount of current is supplying
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solution
as we know
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A diver leaps from the edge of a diving platform
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Explanation:

Below is an attachment containing the solution.

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Question 17 A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off
ivanzaharov [21]

Complete Question:

A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.

Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits

(Question attached)

Answer:

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C (rounded to 1 decimal place)

Explanation:

A calorimeter is used to measure the heat of chemical or physical reactions. The example given in the question is using the calorimeter to determine the specific heat capacity of iron.

When the system reaches equilibrium the iron and water will be the same temperature, T_{e}. The energy lost from the iron will be equal to the energy gained by the water. It is assumed that the only heat exchange is between the iron and water and no exchange with the surroundings.

Q=mc(T_{e}-T_{initial}) (Eq 1)

-Q_{iron}=Q_{water} (Eq 2)

Water:

m_{water}=100.0 g, c_{water}=4.186 J/kg.\°C, T_{initial,water}=23 \°C, T_{e}=27.6 \°C

Iron:

m_{iron}=59.1 g, c_{iron} = ? J/kg.\°C, T_{initial,iron}=85 \°C, T_{e}=27.6 \°C

Substituting Eq 1 into Eq 2 and details extracted from the question:

-m_{iron}c_{iron}(T_{iron,e}-T_{initial})=m_{water}c_{water}(T_{water,e}-T_{initial})

-59.1*c_{iron}(27.6-85)=100.0*4.186(27.6-23)

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C

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They are relatively straight or as straight as they are going to get during a given month that is when a full moon occurs.
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