In a running event, a sprinter does 4.9 105 J of work and her internal energy decreases by 9.1 105 J.
(a) Determine the heat transferred between her body and surroundings during this event.
J
(b) What does the sign of your answer to part (a) indicate?
There is no heat transferred between the sprinter and the environment. Energy is transferred from the environment to the sprinter by heat. Energy is transferred from the sprinter to the environment by heat.
To determine the heat transferred between the sprinter's body and surroundings during the event, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the sum of the heat transferred to or from the system and the work done on or by the system.
The equation for the first law of thermodynamics is:
ΔU = Q - W
Where:
ΔU is the change in internal energy
Q is the heat transferred
W is the work done
Given that the sprinter's internal energy decreases by 9.1 x 10^5 J (negative value) and she does 4.9 x 10^5 J of work (positive value), we can substitute these values into the equation:
-9.1 x 10^5 J = Q - 4.9 x 10^5 J
Simplifying the equation, we can isolate Q:
Q = -9.1 x 10^5 J + 4.9 x 10^5 J
= -4.2 x 10^5 J
Therefore, the heat transferred between the sprinter's body and surroundings during the event is -4.2 x 10^5 J.
Now let's move on to part (b) to understand the sign of the answer.
The negative sign indicates that heat is transferred from the sprinter to the surroundings. In other words, the sprinter loses heat energy to the environment. This means that the sprinter is releasing thermal energy to the surroundings, likely due to the exertion and increased metabolism during the running event.
To determine the heat transferred between the body and surroundings during the event, we need to use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
Given that the sprinter's internal energy decreases by 9.1 10^5 J (ΔU = -9.1 10^5 J) and the work done by the sprinter is 4.9 10^5 J (W = 4.9 10^5 J), we can plug these values into the equation to find the heat transferred (Q):
-9.1 10^5 J = Q - 4.9 10^5 J
Rearranging the equation to solve for Q, we have:
Q = -9.1 10^5 J + 4.9 10^5 J
Q = -4.2 10^5 J
Therefore, the heat transferred between the sprinter's body and the surroundings during the event is -4.2 10^5 J.