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2 years ago

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A large pot is placed on a stove and 1.2 kg of water at 14°C is added to the pot. The temperature of the water is raised evenly

to 100°C just before it starts to boil. (a) What amount of heat is absorbed by the water in reaching 100°C? (b) The water then boils until all of it has evaporated, turning to water vapor at 100°C. How much heat does the water absorb in this process?

1 answer:

geniusboy [140]2 years ago

4 0

Answer:

a) the amount of heat absorbed by the water is 431995.2 J

b) the amount of heat absorbed during evaporation is 2712000 J

Explanation:

Given that;

mass of water Mw = 1.2 kg

Specific heat capacity of water Cw = 4186 J/kg.C

Change in temperature ΔT = final T - Initial T = 100 - 14 = 86°C

Now

A)

Heat required to raise the temperature of water is expressed as:

Q = Mw × Cw × ΔT

Q = 1.2 × 4186 × 86

Q = 431995.2 J

Therefore the amount of heat absorbed by the water is 431995.2 J

B)

Then heat absorbed during evaporation will be:

Q1 = Heat absorbed during phase change from water to steam = Mw × Lv

Lv = latent heat of vaporization of water = 2.26 × 10⁶ J/kg

so

Q1 = 1.2 × 2.26 × 10⁶ = 2712000 J

Therefore the amount of heat absorbed during evaporation is 2712000 J

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A bimetallic strip (brass/steel), which is straight at room temperature, will be immersed in boiling water and allowed to equili

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The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm

given paramers

* Bimetallic brass / steel tape

* Initial temperature, room temperature T = 20ºC

* Final temperature, boiling water = 100ºC

* initial length L₀ = 222mm (1cm / 10mm) = 22.2cm

* thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm

to find

* perpendicular deviation or deflection (Δy)

Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by

ΔL = α L₀ ΔT

ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.

In this case we have a strip formed by two materials with different coefficient of thermal expansion,

Brass α_{brass} = 19 10⁻⁶ ºC⁻¹

Steel α_{steel} = 11 10⁻⁶ ºC⁻¹

In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less

thermal expansion. Let's find the length of the strip for each material

brass L_{f brass} - L₀ = α_{brass} L₀ ΔT

Steel L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT

Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.

If we assume that we have an arc of circumference, we can write the length of the arc

θ = L / r

where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material

brass L_{f \ brass} =θ r₁

steel L_{f \ steel} = θ r₂

we substitute in our equations

θ r₁ - L₀ = α_{brass} L₀ ΔT

θ r₂ - L₀ = α_{steel} L₀ ΔT

let's subtract the two equations

θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})

the thickness of the strip is

e = r₁ -r₂

θ = $Lo \ \Delta T \ \frac{\alpha_{brass} - \alpha_{steel}}{e}$

we calculate

θ = $22.2 \ (100-20) \ \frac{(19-11) \ 10^{-6}}{0.0914}$

θ = 0.155 rad

Let's use trigonometry to find the perpendicular deflection

tan θ = Δy / L₀

Δy = L₀ tan θ

Δy = 22.2 tan 0.155

Δy = 3.48 cm

Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm

Learn more about thermal expansion here: brainly.com/question/18717902

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Answer:

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