A helium-filled balloon has a volume of 2.75 L at 20.°C. The volume of the
balloon decreases to 2.46 L after it is placed outside on a cold day. What is the
outside temperature in K? in °C?

The outside temperature is 262.1K or -10.9°C

Explanation:

Given that,

Original volume V1 = 2.75 L

Original temperature T1 = 20.0°C

Since unit of temperature is Kelvin, convert 20°C to Kelvin by adding 273 to 20°C

(20.0°C + 273K = 293K)

Outside volume V2 = 2.46 L

Outside temperature T2 = ?

Since volume and temperature are given while pressure is constant, apply the formula for Charle's law

V1/T1 = V2/T2

2.75L/293K = 2.46L/T2

To get the value of T2, cross multiply

2.75L x T2 = 2.46L x 293K

2.75LT2 = 720.78LK

Divide both sides by 2.75L

2.75LT2/ 2.75L = 720.78LK/2.75L

T2 = 262.1K

Hence, the outside temperature in Kelvin is 262.1K. Then, obtain outside temperature in Celsius by subtracting 273 from 262.1K

i.e 262.1K - 273 = -10.9°C

Thus, the outside temperature is 262.1K, or -10.9°C

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.