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?

317.46 °C

Explanation:

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

Where,  

 is the heat absorbed/released

m is the mass

C is the specific heat capacity

 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

Q = 4700 J

So,  

Thus, the final temperature is:- 317.46 °C

answer carbon and three oxygen atoms;