A 41 g Ice cube at -21 C is dropped into a container of water at 0 C. How much water freezes onto the ice? The specific heat of ice is .5 cal/g C and it's heat of fusion of is 80 cal/g.

Answers

Answer 1

Answer:

The right solution is "5.38 grams".

Explanation:

The given values are:

Heat of fusion,

L = 80 cal/g

Mass of ice cube,

[tex]m_{ice} = 41 \ g[/tex]

Specific heat of ice,

[tex]C_{ice}=0.5 \ cal/g[/tex]

Let,

Gram of water freezes will be "m".

⇒  [tex]mL=m_{ice} C_{ice} (0+21)[/tex]

Or,

⇒     [tex]m=\frac{m_{ice} C_{ice} (0+21)}{L}[/tex]

On substituting the values, we get

⇒         [tex]=\frac{41\times 0.5\times 21}{80}[/tex]

⇒         [tex]=\frac{430.5}{80}[/tex]

⇒         [tex]=5.38 \ grams[/tex]


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Answers

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Answer:

the answer to this question will be False

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false.......,.....,...........

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Answers

1/2 m/s2
30 Na




I hope this helps!

We have that  the Time  it took him to reach that speed is

t=1.2sec

From the question we are told

A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s. If the runner has a mass of  60 kg, how long did it take him to reach that speed?

Generally the equation for the   is mathematically given as

[tex]P=W/t[/tex]

Therefore

[tex]W=Pt\\\\W=250*t[/tex]

Where

[tex]W=f*d\\\\5*60=250*t\\\\t=\frac{5*60}{250}]\\\\t=1.2sec[/tex]

Therefore

Time  it took him to reach that speed is

t=1.2sec

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