Question

Topic: Calorimetry

Question:
Mr. Roosevelt wants to boil water by shaking it in a flask. Suppose 1.102 lbs of water at 20℃ falls 0.0795 rod in each shake. If he makes 30 such shakes per minute. Neglecting any heat loss calculate the time taken by Mr. Roosevelt to boil the water.

"Boiling" here means that water has to be boiled at the temperature-100℃ (vapours). ​

Answer 1

1.102 lbs = 1.102 × 0.45 = 0.5 kg

0.0795 rod = 0.0795 × 5.03 = 0.4 m

Now, it is given that we have to convert 0.5kg of water(at 20℃) into vapours at 100℃.

Energy used in shaking the flask = mC∆T + mL

=> m×g×h = mC∆T + mL

Thus, Work done in shaking the flask for time 't' is: (for 30 such shakes)

=> 30(mgh) = (mC∆t + mL)/t

=> 30×0.5×10×0.4 = (0.5×4200×80 + 0.5×2156)/t

=> 30 = (168000 + 1078)/t

=> 30 = 169078/t

=> t = 169078/30

=> t = 5635.93 minutes

=> t = (5635.95)/(24×60) days

=> t = 3.91 days

or

=> t = 3-days 22-hours and 24 minutes

Hope it helps you william bro!

In case of any query in my solution please ask me.

Answer 2

⭐ SOLUTION ⭐

$\implies \: energy \: used \: to \: shake \: the \: flask \: = mc \triangle \: t \: + ml$

So,

$\implies \: mgh = mc∆t + mL$

➪NOTE:- Here , 1.102lbs = 0.5kg.(1.102×0.45)&

And , 0.0795 = 0.4m(0.0795×5.03)

➪ Work done in shaking the flask will be ,

$\implies \: mgh = \frac{mc \triangle \: t + ml}{t}$

$\implies \: 0.5 \times 10 \times 0.4 = \frac{(0.5 \times 4200 \times 80 + 0.5 \times 2156)}{t}$

$\implies \: 30 = \frac{ 168000 + 1078}{t}$

$\implies \: 30 = 169078 \div t \\ \\ \implies \: t = 5635.93min.$

Which is ,

$\implies \: \frac{5635.95}{24 \times 60}$ days

$\implies \: t = 3.91days.$

Hence, time taken by Mr.roosevelt to boil the water is ,

$\boxed{ \implies \: \: t = 3.91days}$

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HOPE IT HELPS :)