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

Answer:

317.46 °C

Explanation:

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

$Q=m\times C\times \Delta T$

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,  

$4700=118.5\times 0.129\times (x-10)$

$118.5\times \:0.129\left(x-10\right)=4700$

$x-10=307.46083$

$x=317.46$

Thus, the final temperature is:- 317.46 °C