Question

118.5 g piece of lead is heated with 4,700 J of energy. If the specific heat of lead is 0.129 J/ (g ⋅ °C), and the lead’s initial temperature was 10 °C, what is the final temperature of the lead?

Answer 1

317.46 °C

Step-by-step explanation:

The expression for the calculation of heat is shown below as:-

`Q=m* C* \Delta T`

Where,

`Q` is the heat absorbed/released

m is the mass

C is the specific heat capacity

`\Delta T` is the temperature change

Thus, given that:-

Mass of lead = 118.5 g

Specific heat = 0.129 J/g°C

Initial temperature = 10 °C

Final temperature = x °C

`\Delta T=(x-10)\ ^0C`

Q = 4700 J

So,

`4700=118.5* 0.129* (x-10)`

`118.5* \:0.129\left(x-10\right)=4700`

`x-10=307.46083`

`x=317.46`

Thus, the final temperature is:- 317.46 °C