In: Chemistry
An ice chest at a beach party contains 12 cans of soda at 2.45 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 7.45-kg watermelon at 26.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.
Heat lost by watermelon = heat gained by soda.
Let the equilibrium temperature of soda-watermelon = XoC.
Specific heat of watermelon = 4.186J/g.oC
Mass of watermelon = 7.45kg = 7450g
Initial temperature of watermelon = 26.4oC
∆T for watermelon = 26.4-X (As hotter body loses temperature)
Mass of each can of soda = 0.35kg
Total cans = 12
Total mass of soda = 12 × 0.35kg = 4.2kg
Specific heat of soda = 3800J/kg.oC
Initial temperature of soda = 2.45oC
∆T for soda = X-2.45 (As cooler body gains temperature)
So
7450g × 4.186 J/g.oC × (26.4-X)oC = 4.2kg × 3800 J/kg.oC × (X-2.45)oC
On solving X = 18.3oC.This is the final temperature of soda and watermelon.