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Vinil7 [7]
3 years ago
15

An unknown mass of each substance, initially at 25.0 ∘C, absorbs 1920 J of heat. The final temperature is recorded. Find the mas

s of each substance. You may want to reference (Pages 259 - 264) Section 6.4 while completing this problem.
Physics
1 answer:
zhannawk [14.2K]3 years ago
7 0

Answer: mass for Pyrex glass 84.21g

mass for sand 61.6g

mass for ethanol 41.32g

mass for water 62.07g

Explanation

By definition specific heat is the amount of heat required to change the temperature of 1 kg mas by 1°C

Q=mcΔT is formula for specific heat

Q is heat transfer

m is mass

ΔT is change in temperature

c   is specific heat

c of Pyrex glass= 0.75 j/g°C

c of sand = 0.84 j/g°C

c of ethanol= 2.42 j/g°C

c of water = 4.18 j/g°C

now we will make M(mass) the subject, so equation becomes

m=Q/cΔT

for

pyrex glass T<em>f=</em>55.4°C

m=1920/(55.4-25)*0.75

m=84.21g {after cutting J(joules) and °C we are left with g(grams)}

for

sand T<em>f</em>=62.1°C

m=1920/(62.1-25)*0.84

m=61.6g {after cutting J(joules) and °C we are left with g(grams)}

for

ethanol T<em>f</em>=44.2°C

m=1920/(44.2-25)*2.42

m=41.32g  {after cutting J(joules) and °C we are left with g(grams)}

for

water T<em>f=</em>32.4°

m=1920/(32.4-25)*4.18

m=62.07g  {after cutting J(joules) and °C we are left with g(grams)}

i hope you understand the solution, thank you.

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Answer:

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40 seconds

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Explanation:

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Differentiating with respect to time we get

\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s

b) The object will reach the highest point after 20 seconds

s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m

c) Highest point the object will reach is 360 m

s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0

\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s

d) Time taken to strike the ground would be 20+20 = 40 seconds

[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s

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e) The velocity with which the object strikes the ground will be 36 m/s

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A 10.0kg water balloon is dropped from a height of 12.0m. Calculate the speed of the balloon just before it hits the ground
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15.5 m/s.

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