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3 years ago

NEED HELP NOW PLEASE! NEED ANSWER AND EXPLANATION!!!

1 answer:

6 0

The first law of thermodynamics states the conversation of energy and heat where the energy in an isolated system may be transformed into another, but never created or destroyed. If 314J of energy was released to the room, then also 314J of energy was removed from food in that refrigerator assuming it is an isolated system

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Which of the following lists contain only elements?

N76 [4]

In which elements are belonging to? I will grant that option B ; ti's the best answer.
In which thou might contain the elements zinc , gold , aluminium , and last by not least oxygen.

I truly hope ti's answer helps thou.

7 0

3 years ago

3. How can you decrease the pressure of a gas in a container without changing the volume of the gas?

inna [77]

Answer:

reduce the temperature of the gas

Explanation:

when you reduce the temperature of the gas the pressure will decrease

8 0

3 years ago

Determine the volume, in liters, occupied by 0.015 molecules of oxygen at STp?

Arlecino [84]

5.58 X $10^{-25}$ Litres is the volume, in liters, occupied by 0.015 molecules of oxygen at STP.

Explanation:

Data given:

molecules of oxygen = 0.015

number of moles of oxygen =?

temperature at STP = 273 K

Pressure at STP = 1 atm

volume = ?

R (gas constant) = 0.08201 L atm/mole K

to convert molecules to moles,

number of moles = $\frac{molecules}{Avagadro's number}$

number of moles = 2.49 x $10^{-26}$

Applying the ideal gas law since the oxygen is at STP,

PV = nRT

rearranging the equation:

V = $\frac{nRT}{P}$

putting the values in the rearranged equation:

V = $\frac{2.49 X10^{-26} X 0.08201 X 273}{1}$

V = 5.58 X $10^{-25}$ Litres.

5 0

3 years ago

You wish to cool a 1.59 kg block of tin initially at 88.0°C to a temperature of 49.0°C by placing it in a container of kerosene

rjkz [21]

Answer:

We need 0.482 L of kerosene

Explanation:

Step 1: Data given

Mass of the block tin = 1.59 kg

Initial temperature tin= 88.0 °C

Final temperature = 49.0 °C

Initial temperature of kerosene = 32.0 °C

Density of kerosene = 820 kg/m³

Specific heat of tin is 218 J/(kg · °C)

Step 2: Calculate mass of kerosene

Heat lost = heat gained

Qtin = -Qkerosene

Q = m*c*ΔT

m(tin) * c(tin) * ΔT(tin) = -m(kerosene) * c(kerosene) * ΔT(kerosene)

⇒ mass of tine = 1.59 kg

⇒ c(tin) = the specific heat of tin = 218 J/ kg*°C

⇒ ΔT(tin) = The change in temperature = T2 - T1 = 49.0 °C - 88.0 °C = -39.0 °C

⇒ mass of kerosene = TO BE DETERMINED

⇒ c(kerosene) = The specific heat of kerosene = 2010 J/kg*°C

⇒ ΔT = 49.0 - 32.0 = 17.0

1.59 kg * 218 J/kg*°C * -39.0 °C = - m(kerosene) * 2010 J/kg*°C *17.0 °C

-13518.18 = -m(kerosene) * 34170

m(kerosene) = 0.39562 kg

Step 3: Calculate volume of kerosene

Volume = mass / density

Volume = 0.39562 kg / 820 kg/m³

Volume = 4.82 * 10^-4 m³ = 0.482 L

We need 0.482 L of kerosene

6 0

3 years ago

What type of temperatures and moisture content does Marine time Tropical air have?

Alex777 [14]

Warm temps/ Moist air

4 0

3 years ago