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 initia l temperature was 10 °C, what is the final temperature of the lead?
1 answer:
Answer:
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,
<u>Thus, the final temperature is:- 317.46 °C </u>
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