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3 years ago

Explain why energy from the sun cannot reach us by conduction or convection?

Physics

2 answers:

34kurt3 years ago

8 0

Answer:

It cannot reach us by conduction or convection as there is no medium such as air in most part of the space between the earth and the sun. From the sun the heat comes to us by another process known as radiation. The transfer of heat by radiation does not require any medium.

k0ka [10]3 years ago

8 0

Answer:

Heat can't be transferred from sun to earth by conduction, because air is a bad conductor of heat. Convection is the transfer of heat in gases & liquids. Convection occurs when a heated fluid moves away from the source of heat and comes in contact with other substances, transferring some of their energy.

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

Step 4: calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

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3 years ago

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Zinaida [17]

Answer:

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Explanation:

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2 years ago

Determine the elastic energy U stored in the compressed spring.

hoa [83]

Answer:

Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.

Explanation:

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2 years ago

What is silviculture?

expeople1 [14]

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3 years ago

If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?

HACTEHA [7]

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