In a game of football outdoors on a cold day, a player will begin to feel exhausted after using approximately 8.0 Ă— 105 J of internal energy. (a) One player, dressed too lightly for the weather, has to leave the game after losing 6.8 Ă— 105 J of heat. How much work has he done? (b) Another player, wearing clothes that offer better protection against heat loss, is able to remain in the game long enough to do 2.4 Ă— 105 J of work. What is the magnitude of the heat that he has lost?

Respuesta :

Answer:

Part a)

[tex]W = 1.2 \times 10^5 J[/tex]

Part b)

[tex]Q = 5.6 \times 10^5 J[/tex]

Explanation:

It given that player will feel exhausted when he is using his internal energy of [tex]8.0 \times 10^5 J[/tex]

PART a)

it is given that heat loss by the player is given as

[tex]Q = 6.8 \times 10^5 J[/tex]

now by first law of thermodynamics we have

[tex]\Delta U = Q + W[/tex]

now we have

[tex]8.0 \times 10^5 = 6.8 \times 10^5 + W[/tex]

[tex]W = 1.2 \times 10^5 J[/tex]

PART b)

It is given that another player did the work as

[tex]W = 2.4 \times 10^5 J[/tex]

now we have first law of thermodynamics

[tex]\Delta U = Q + W[/tex]

now we have

[tex]8.0 \times 10^5 = 2.4 \times 10^5 + Q[/tex]

[tex]Q = 5.6 \times 10^5 J[/tex]