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harkovskaia [24]
3 years ago
11

A 1 m ID spherical cooler made of 2 cm thick foam insulation (k = 0.15 Wm-1K-1) is used to store iced water at 0°C. The tank is

located outdoors at 25 °C and is subjected to horizontal cross winds at 1 m/s. Determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0 °C that melts during a 24 h period. Assume the inner surface of the cooler is at 0 °C and take the thermal resistance of the cooler and heat transfer by radiation into consideration. Assume the average surrounding surface temperature for radiation exchange to be 20 °C and the outer surface of the cooler to have an emissivity of 0.85. The heat of fusion of water at atmospheric pressure is hif = 333.7 kJ/kg.
Chemistry
1 answer:
Leto [7]3 years ago
3 0

The answer is unknown

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what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c
slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

Q = m \times C \times \bigtriangleup T

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

C = \frac{Q}{m \times \bigtriangleup T}

Let us plug in the values in above equation.

C = \frac{300J}{267g \times 12 C}

C = \frac{300J}{3204 g C}

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

3 0
3 years ago
The question is in the attached word document.
Alekssandra [29.7K]

Answer:i read it Explanation:

3 0
2 years ago
Which two compounds are classified as bases by the Brønsted-Lowry definition, but not by the Arrhenius definition, and why?
konstantin123 [22]

Answer: Ammonia (NH3) and sodium carbonate (Na2CO3), because they accept hydrogen ions but lack hydroxide ions.

Explanation:

i took the test and got it correct :) hope this helps

6 0
3 years ago
Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
3 years ago
3. A mass of 0.15 ounces is equal to how many grams?
garik1379 [7]

Answer:

Option C = 4.25 g

Explanation:

Ounce and grams are unit of mass. Ounce is larger unit while gram is smaller unit. The one ounce is consist of 28.35 g or we can say that one ounce is equal to 28.35 g. In order to convert the given ounce value into grams the value is multiply with 28.35 g.

Given data:

Mass = 0.15 ounce

Mass in gram = ?

Solution:

One ounce is equal to 28.35 g, so

0.15 × 28.35 = 4.25 g

5 0
3 years ago
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