Question

A piece of copper metal is initially at 100.0°C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a temperature of 20.0°C. After stirring, the final temperature of both copper and water is 25.0°C. Assuming no heat losses, and that the specific heat (capacity) of water is 4.18 J/(g·K), what is the heat capacity of the copper in J/K?

Answer 1

C = 13.9 J/K

Step-by-step explanation:

Here we are given that no energy is lost in the surrounding so we can say that there energy is conserved in this system

So we will have

energy given by hot piece of copper = energy absorbed by the coffee

So we will have

`m_1c_1\Delta T_1 = m_2c_2\Delta T_2`

so we will have

`m_2 = 50 g`

`C_2 = 4.18 J/g K`

`\Delta T_1 = 100 - 25`

`\Delta T_2 = 25 - 20`

`C(100 - 25) = 50(4.18)(25 - 20)`

`75 C = 1045`

`C = 13.9 J/K`