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Digiron [165]
3 years ago
6

State the difference between sensible heat and latent heat

Physics
1 answer:
Anna11 [10]3 years ago
6 0

<u>Latent hea</u>t is related to changes in phase between liquids ,gases and solids.

<u>Sensible heat</u> is related to changes in temperature of a gas or object with no change in phase.

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Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m
Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:  

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12        ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4        ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

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3 years ago
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Nuclear fusion produces elements that are heavier than helium.

Explanation:

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What is needed for a complete circuit? a Charge flows along a complete conducting path b High Resistance c Wires connected to a
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a Charge flows along a complete conducting path

Explanation:

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6) 100 ml of water is initially at 20°C. 30,000 J of heat is added to the water. What is temperature change for the water?
vovangra [49]
<h2>Δt = 71.67 °C</h2>

The temperature change of water is equal to 71.67 °C

<h3>Explanation:</h3>

Given:

Amount of transferred energy = 30,000 K J

Mass of water = 100 ml

Initial temperature = 20°C

To find the change in temperature of water.

Formula for Heat capacity is given by

Q = m×c×Δt ........................................(1)

where:

Q = Heat capacity of the substance (in J)

m=mass of the substance being heated in grams(g)

c = the specific heat of the substance in J/(g.°C)

Δt = Change in temperature (in °C)

Δt = (Final temperature - Initial temperature) = T(f) - T(i)

Q = 30,000 J

Mass of water = m = 100 ml

1 ml = 1 g ................................................(2)

Therefore m = 100 ml = 100 g

Specific heat of water is c = 4.186 J /g.

Δt = ?

Substituting these in equation (1), we get

Q = m×c×Δt

Rearranging the terms for Δt,

Δt = \frac{Q}{m\times c}

Δt = \frac{30,000}{100\times 4.186}  = \frac{30,000}{418.6}= 71.67\°C

Δt = 71.67 °C

8 0
3 years ago
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