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3 years ago

13

A 214 g aluminum cup holds and is in thermal equilibrium with 892 g of water at 76°C. The combination of cup and water is cooled

uniformly so that the temperature decreases by 0.9°C per minute. At what rate is energy being removed?

2 answers:

givi [52]3 years ago

8 0

Answer: rate at which energy is being removed is 58.9J/s or 58.9W

Explanation:

Given;

Mass of water Mw =892g = 0.892kg

Mass of aluminium Ma = 214g = 0.214kg

Rate of change in temperature ∆T = 0.9°C per minute

Specific heat capacity of water Cw = 4186J/kgC

Specific heat capacity of aluminium Ca = 900J/kgC

Since both the water and aluminium are at thermal equilibrium and the decrease with the same rate, the energy removed per minute is given as;

Q = (MwCw + MaCa)∆T

Q = (0.892×4186 + 0.214×900)×0.9

Q = 3533.8608J per minute

To determine the energy being removed in watts

P = Q/t

P = 3533.8608J/60s

P = 58.89768W

P = 58.9W

gladu [14]3 years ago

7 0

Answer:

<em>The rate at which energy is removed = 59.085 J</em>

Explanation:

The rate at which energy is removed = c₁m₁ΔΘ/t + c₂m₂ΔΘ/t................. Equation 1

Where c₁ = specific heat capacity of the aluminum, m₁ = mass of the aluminum, ΔΘ/t = rate of temperature decrease, c₂ = specific heat capacity of water, m₂ = mass of water.

<em>Given: m₁ = 214 g = (214/1000) kg = 0.214 kg, m₂ = 892 g = (892/1000) kg </em>

<em>m₂ = 0.892 kg, ΔΘ/t = 0.9 °C/min. = (0.9/60) °C/seconds. = 0.015 °C/seconds.</em>

<em>Constants: c₁ = 900 J/kg.°C, c₂ = 4200 J/kg.°C</em>

<em>Substituting these values into Equation 1,</em>

<em>The rate at which Energy is removed = (0.214×900×0.015) + (0.892×4200×0.015)</em>

<em>The rate at which energy is removed = 2.889 + 56.196</em>

<em>The rate at which energy is removed = 59.085 J</em>

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Answer with Explanation:

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3 years ago

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Answer:

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Explanation:

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Angle = 74.9°

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We need to calculate the time

Using formula of time

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We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

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Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago