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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?

Respuesta :

Answer:

317.46 °C

Explanation:

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

[tex]Q=m\times C\times \Delta T[/tex]

Where,  

[tex]Q[/tex]  is the heat absorbed/released

m is the mass

C is the specific heat capacity

[tex]\Delta T[/tex]  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

[tex]\Delta T=(x-10)\ ^0C[/tex]

Q = 4700 J

So,  

[tex]4700=118.5\times 0.129\times (x-10)[/tex]

[tex]118.5\times \:0.129\left(x-10\right)=4700[/tex]

[tex]x-10=307.46083[/tex]

[tex]x=317.46[/tex]

Thus, the final temperature is:- 317.46 °C

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