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A sample of solid platinum is heated with an electrical coil. If 30.0 Joules of energy are added to a 13.1 gram sample and the final temperature is 38.1 °C, what is the initial temperature of the platinum?

Respuesta :

Answer:

[tex] Q = m C_p \Delta T = m C_p (T_f -T_i)[/tex]

And we have the following info given:

[tex] m = 13.1 gr *\frac{1Kg}{1000 gr}= 0.0131 Kg[/tex]

[tex] T_f = 38.1 C[/tex] represent the initial temperature

[tex] Q = 30 J[/tex] represent the heat added

From tables for the platinum we know:

[tex] C_p = 150 \frac{J}{Kg K}[/tex]

And if we solve for the initial temperature we got:

[tex] \frac{Q}{m C_p} = T_f -T_i[/tex]

[tex] T_i = T_f - \frac{Q}{m C_p}[/tex]

And replacing we got:

[tex] T_f = 38.1C -\frac{30 J}{0.0131 Kg * 150 \frac{J}{Kg C}}= 22.833 C[/tex]

Explanation:

For this case we can assume that the only mechanism of heat transfer is the associated to the sensible heat given by this formula:

[tex] Q = m C_p \Delta T = m C_p (T_f -T_i)[/tex]

And we have the following info given:

[tex] m = 13.1 gr *\frac{1Kg}{1000 gr}= 0.0131 Kg[/tex]

[tex] T_f = 38.1 C[/tex] represent the initial temperature

[tex] Q = 30 J[/tex] represent the heat added

From tables for the platinum we know:

[tex] C_p = 150 \frac{J}{Kg K}[/tex]

And if we solve for the initial temperature we got:

[tex] \frac{Q}{m C_p} = T_f -T_i[/tex]

[tex] T_i = T_f - \frac{Q}{m C_p}[/tex]

And replacing we got:

[tex] T_f = 38.1C -\frac{30 J}{0.0131 Kg * 150 \frac{J}{Kg C}}= 22.833 C[/tex]

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