Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volume was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer?

Explanation:

Given:

Initial volume of the balloon V1 = 348 mL

Initial temperature of the balloon T1 = 255C

Final volume of the balloon V2 = 322 mL

Final temperature of the balloon T2 =

To calculate T1 in kelvin

T1= 25+273=298K

Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula

T2=( V2*T1)/V1

T2=(322*298)/348

Hence, the temperature of the freezer is 276 K

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.

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