In a game of football outdoors on a cold day, a player will begin to feel exhausted after using approximately 8.0×105j of internal energy. (a) one player, dressed too lightly for the weather, has to leave the game after losing 6.8×105j 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 2.1×105j of work. what is the magnitude of the heat that he has lost?

A work of was done by the first player.

The quantity of heat lost by the second player is

Explanation:

From First Law of Thermodynamics,

ΔQ - W = ΔU

(a) Internal Energy used up by a player, ΔU = -8.0 X 10⁵ J

The player lost 6.8 X 10⁵ J of Heat (ΔQ)

∴ W = ΔQ - ΔU ............... (1)

W = -(6.8 X 10⁵) - (-8.0 X 10⁵)

W = -6.8 X 10⁵ + 8.0 X 10⁵

W = 1.2 X 10⁵ J ( Work done by the player).

(b) Heat lost by the second player

W = 2.1 X 10⁵ J;  ΔU = -8.0 X 10⁵ J

ΔQ = W + ΔU  (re-arranging equation 1)

ΔQ = 2.1 X 10⁵ + (-8.0 X 10⁵)

ΔQ = 2.1 X 10⁵ - 8.0 X 10⁵

∴ ΔQ = -5,9 X 10⁵ J (Heat lost by the second player).

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation