Answer to Question #87385 in Chemistry for carter

Question

a 12 oz can of soda weighs about 450.0 g how many joules are released when a can of soda is cooled from 25 c to 4 c. which c constant should be used for this calculation

Answer 1

According to the heat equation:

$$Q = c × m × (Tf - Ti)$$

where Q is a heat, c is a specific heat of the object, m is a mass, Tf is a final temperature, Ti is an initial temperature.

As most of the object (can of soda) consists of water, the heat capacity of water should be used: c = 4.18 J/g×°C.

$$Q = 4.18 J/g×°C × 450 g × (4°C - 25°C) = -39.5 kJ$$

Answer: 39.5 kJ of heat are released.