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Korvikt [17]
2 years ago
14

In an experiment to measure the specific heat capacity of copper, 0.02 kg of water at 70°C is poured into a copper calorimeter (

with a stirrer) of mass 0.16 kg at 15°C. After stirring, the final temperature is found to be 45°C. If the specific heat of water is 4,200 J/kg/°C,​
Physics
1 answer:
harkovskaia [24]2 years ago
6 0

\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTIONS}}}}}}}

In an experiment to measure the specific heat capacity of copper, 0.02 kg of water at 70°C is poured into a copper calorimeter (with a stirrer) of mass 0.16 kg at 15°C. After stirring, the final temperature is found to be 45°C. If the specific heat of water is 4,200 J/kg/°C,

\blue{\huge{\red{\boxed{\green{\mathfrak{GIVEN}}}}}}

WATER:-

Mass :- 0.02 kg at 70°C

Specific Heat Of Water is 4,200 J/kg°C,

COPPER:-

MASS:- mass 0.16 kg at 15°C

Temperatures r according to the part !

{\huge{\huge{\bold{\green{To  \: Find :- }}}}}

what is the quantity of heat released per kg of water per 1°C fall in temperature?

Calculate the heat energy released by water in the experiment in cooling from 70°C to 45°C.

Assuming that the heat released by water is entirely used to raise the temperature of calorimeter from 15°C to 45°C) calculate the specific heat capacity of copper.

\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}}}

{\blue{\star{\red{Part  \: 1 :- What \:   is \:  the  \: quantity \:  of heat \:  released \:  per  \: kg \:  of \:  w ater \:  per  \: 1°C \:   fall \:  in \:  temperature }}}}

\orange{Specific \:  \:  Heat}

It is the amount of the heat released by unit mass of the body per 1°C fall in temperature.

It is the amount of the heat absorbed by the unit mass of the body per 1°C rise in temperature

So quantity of heat released per kg of water per 1°C fall in temperature is equal to the specific heat of the water which is 4,200 J/kg°C,

{\red{\star{\blue{Part  \:2 :-\:  Calculate \:  the  \: heat \:  energy \:  released \:  by  \: water \:  in \:  the \:  experiment \:  in \:  cooling \:  from  \: 70°C  \: to \:  45°C. }}}}

MASS OF WATER --> 0.02 kg

INITIAL TEMPERATURE--> 70°C

FINAL TEMPERATURE--> 45°C

CHANGE IN TEMPERATURE--> (45-70)°C =(- 25)°C

SPECIFIC HEAT OF WATER--> 4200 J/ kg °C

Q =  mc\triangle T \\  \\ Q = 0.02 \times 4200 \times ( -25)  \\  \frac{2}{100}  \times 4200 \times( -  25) =Q \\ ( - 50) \times 42 = Q \\ Q = ( - 2100) \: joules

NEGATIVE INDICATES THAT HEAT IS RELEASED BY THE BODY!

IT MEANS COOLING HAS BEEN TAKEN PLACED.

{\red{\star{\green{Part  \:2 :-\:Assuming \:  that \:  the \:  heat \:  released \:  by  \: water}}}} \\  \\  {\green{is  \: entirely \:  used  \: to  \: raise \:  the temp \: of \:  calorimeter  \: from \:  15°C  \: to \:  45°C }} \\  \\{\green{ calculate  \: the  \: specific \:  heat  \: capacity \:  of  \: copper.}}

MASS OF COPPER :- 0.16 kg

INITIAL TEMPERATURE:- 15°C

FINAL TEMPERATURE:- 45°C

CHANGE IN TEMPERATURE--> (45-15)°C = 30°C

AMOUNT OF HEAT RELEASED BY WATER --> 2100 J ( From second part)

Q =  mc\triangle T \\  \\ 2100 = 0.16 \times c \times 30 \\ 70 = 0.16 \times c \\  \\ c =  \frac{7000}{16}  \\ c = 437.5 \:  \frac{J }{kg°C}

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Computer A is 1.41 times faster than the Computer B

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Assume that number of instruction in the program is 1

Clock time  of computer A is CT_{A} =200 ps

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Effective CPI of computer A is CPI_{A} =1.5

Effective CPI of computer B isCPI_{B} =1.7

CPU time of A is

CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec

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CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec

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An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after
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E = kQs/r.r

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 8.0 s, it rotates 35 rad.
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(a) 1.093 rad/s^2

(b) 4.375 rad/s

(c) 8.744 rad/s

(d)  67.845 rad

Explanation:

initial angular velocity, ωo = 0

time, t = 8s

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(a) Let α be the angular acceleration.

Use second equation of motion for rotational motion

\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}

By substituting the values

35 = 0 + 0.5 x α x 8 x 8

α = 1.093 rad/s^2

(b)  The average angular velocity is defined as the ratio of total angular displacement to the total time taken .

Average angular velocity = 35 / 8 = 4.375 rad/s

(c) Let ω be the instantaneous angular velocity at t = 8 s

Use first equation of motion for rotational motion

ω = ωo + αt

ω = 0 + 1.093 x 8 = 8.744 rad/s

(d) Let in next 5 seconds the angular displacement is θ.

\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}

By substituting the values

θ = 8.744 x 5 + 0.5 x 1.093 x 5 x 5

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