Question

A beach ball is inflated to a volume of 25L of air at 15°C. During
the afternoon, the volume increases by 1L. What is the new
temperature outside?​

Answer 1

Answer:

15.6℃

Explanation:

Given:

V1 = 25 L

V2 = 26 L

T1 = 15℃

T2 = ?

Solution:

V1T2 = V2T1

T2= V2T1/V1

= [(26 L)(15℃)] / 25 L

T2 = 15.6℃

Every 1 liter increase in the volume increases the resulting temperature by 0.6.

I hope this helps, Thank you.

Answer 2

Answer:

Temperature2 = 15.6℃

Explanation:

Given:

Volume 1 = 25 L

Volume 2 = 26 L

Temperature 1 = 15℃

To find:

Temperature 2 = ?

Solution:

formula - Volume1 * Temperature2 = Volume2 * Temperature1

Changing sides

Temperature2 = Volume2 *Temperature1 % Volume 1

= [26 * 15] / 25

Temperature2 = 15.6℃

#SPJ2