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katen-ka-za [31]

4 years ago

9

How many joules of heat are needed to raise the temperature of 50.0 g of aluminum from 10°C to 110°C, if the specific heat of al

uminum is 0.95 J/g°C?

1 answer:

dusya [7]4 years ago

5 0

Answer:

Heat required to raise the temperature of the aluminium is 4750 J

Explanation:

As we know that the heat energy required to raise the temperature of the aluminium is given as

$Q = ms\Delta T$

here we know that

m = 50 g

$\Delta T = 110 - 10$

$\Delta T = 100 ^oC$

so we have

$Q = 50(0.95)(100)$

$Q = 4750 J$

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Read 2 more answers

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

a)

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

b)

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

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3 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

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Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

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$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

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$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

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3 years ago

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Answer:

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From conservation of energy

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G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

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Answer:

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