A student eats a dinner rated at 2000 (food) Calories. He wishes to do an equivalent amount of work in the gymnasium by lifting a 50 kg mass. How many times must he raise the mass to expend this amount of energy? Assume that he raises it a distance of 2 m each time and that he regains no energy when it is dropped to the floor. The acceleration due to gravity is 9.8 m/s2 and 1 food Calorie is 103 calories.

A Calorie is a measure of energy that calculated the amount of energy released from the foon it is broken down. The boy can lift the weight up to 879.49 times.

Calories:

It is a measure of energy that calculated the amount of energy released from the foon it is broken down.

Given food Calories,  

Food Calories = 2000 x 103 calories  

Food Calories = 206000 calories   = 861904 J  

The work done by boy can be calculated by the formula,  

Where,

- mass = 50 kg

- gravitational acceleration = 9.8 m/s²

- height = 2 m

Put the values in the formula,

So, boy can lift the weight,

Therefore, the boy can lift the weight up to 879.49 times.

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879.49

Approximately 880 times

Explanation:

2000 (food) Calories = 2000 * 103 calories

2000 (food) Calories = 206000 cal.

Work done by the boy is given by Mgh , potential work

Work= 50*2*9.8

Work = 980 joules.

1 kilocalorie =

4184 joules

So 1 cal = 4.185 joules

206000 calories = 206000*4.184

206000 cal = 861904 joules.

To extend 206000 cal,

He will lift the 50 kg mass 861904/980 times

= 879.49 times.

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask

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explanation: