An ice chest at a beach party contains 12 cans of soda at 2.45 °C. Each...

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.

Solutions

Expert Solution

Heat lost by watermelon = heat gained by soda.

Let the equilibrium temperature of soda-watermelon = X^oC.

Specific heat of watermelon = 4.186J/g.^oC

Mass of watermelon = 7.45kg = 7450g

Initial temperature of watermelon = 26.4^oC

∆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 = 2.45^oC

∆T for soda = X-2.45 (As cooler body gains temperature)

7450g × 4.186 J/g.^oC × (26.4-X)^oC = 4.2kg × 3800 J/kg.^oC × (X-2.45)^oC

On solving X = 18.3^oC.This is the final temperature of soda and watermelon.

queen_honey_blossom answered 1 year ago

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