Physics, 16.03.2020 19:30 ike9264

You heat a 5.2 gram lead ball (specific heat capacity = 0.128 Joules/g-deg) to 183. ∘ C. You then drop the ball into 34.5 milliliters of water (density 1 g/ml; specific heat capacity = 4.184 J/g-deg ) at 22.4 ∘ C. What is the final temperature of the water when the lead and water reach thermal equilibrium?

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Answer from: brandiwingard

The final temperature when the lead and water reach thermal equilibrium is $23.14^{o} C$

Explanation:

Using the law of conservation of energy the heat lost by the lead ball is gained by the water.

$-q_{L} =+q_{w}$ .............................1

where

$q_{L}$ is the heat energy of the lead ball;

$q_{w}$ is the heat energy of water;

but q = msΔ

T...............................2

substituting expression in equation 2 into the equation 1;

T =Δ

T .............................3

where $m_{L}$ is the mass of the lead ball = 5.2 g

$s_{L}$ is the specific heat capacity of the lead ball = 0.128 Joules/g-deg

$m_{w}$ is the mass of water = volume x density = 34.5 ml x 1 g/ml = 34.5 g

$s_{w}$ is the specific heat capacity of water = 4.184 J/g-deg

T is the temperature changes encountered by the lead ball and water respectively

inputting the values of the parameters into equation 3

5.2 g x 0.128 Joules/g-deg x (183-22.4) = 34.5 g x 4.184 J/g-deg x ( $T_{2}$ -22.4)

$T_{2}$ -22.4 = $\frac{106.9}{144.3}$

$T_{2}$ -22.4 = 0.7408

$T_{2}$ = $23.14^{o} C$

Therefore the final temperature when the lead and water reach thermal equilibrium is $23.14^{o} C$

Answer from: Quest

F= (mass)(acceleration) = mam = 0.25 kgvi = 16 m/st = 2 svf = 0 m/s (since it was put to stop)a=(vf-vi)/ta=(0-16)/2a = 8 m/s^2 (decelerating)f = ma = (0.25 kg)(8 m/s^2)f = 2 nhope this !

Answer from: Quest

Stars are born within the clouds of dust and scattered throughout most galaxies. a familiar example of such as a dust cloud is the orion nebula. turbulence deep within these clouds gives rise to knots with sufficient mass that the gas and dust can begin to collapse under its own gravitational attraction. as the cloud collapses, the material at the center begins to heat up. known as a protostar, it is this hot core at the heart of the collapsing cloud that will one day become a star. three-dimensional computer models of star formation predict that the spinning clouds of collapsing gas and dust may break up into two or three blobs; this would explain why the majority the stars in the milky way are paired or in groups of multiple stars.therefor, i would say the answer is: dust and gases contract.i hope this ! < 3

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You heat a 5.2 gram lead ball (specific heat capacity = 0.128 Joules/g-deg) to 183. ∘ C. You then dr...