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Bond [772]
3 years ago
10

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

er containing 75 g of water at 20°C.
The calorimeter is constructed of a material that has a specific heat of 0.10 cal/ g⋅°C.

When equilibrium is reached, what will be the final temperature? cwater = 1.00 cal/g⋅°C.

a. 114°Cb. 72°Cc. 64°Cd. 37°C
Physics
1 answer:
Gnom [1K]3 years ago
6 0

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

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Three joggers are running along straight lines as follows: Jogger A, with a mass of 55.2kg, is traveling along the line y=6.00m
frutty [35]

Answer:

L = - 1361.591 k Kgm/s

Explanation:

Given

mA = 55.2 Kg

vA = 3.45 m/s

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vB = 4.23 m/s

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mC = 72.1 Kg

vC = 4.75 m/s

rC = - 5.00 m

then we apply the equation

L =  (mv x r)

⇒  LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s

⇒  LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s

⇒  LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s

Finally, the total counterclockwise angular momentum of the three joggers about the origin is

L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k  Kgm/s

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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
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Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

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