Question

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?

Answer 1

`Q = m C_p \Delta T = m C_p (T_f -T_i)`

And we have the following info given:

`m = 13.1 gr *(1Kg)/(1000 gr)= 0.0131 Kg`

`T_f = 38.1 C` represent the initial temperature

`Q = 30 J` represent the heat added

From tables for the platinum we know:

`C_p = 150 (J)/(Kg K)`

And if we solve for the initial temperature we got:

`(Q)/(m C_p) = T_f -T_i`

`T_i = T_f - (Q)/(m C_p)`

And replacing we got:

`T_f = 38.1C -(30 J)/(0.0131 Kg * 150 (J)/(Kg C))= 22.833 C`

Step-by-step 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:

`Q = m C_p \Delta T = m C_p (T_f -T_i)`

And we have the following info given:

`m = 13.1 gr *(1Kg)/(1000 gr)= 0.0131 Kg`

`T_f = 38.1 C` represent the initial temperature

`Q = 30 J` represent the heat added

From tables for the platinum we know:

`C_p = 150 (J)/(Kg K)`

And if we solve for the initial temperature we got:

`(Q)/(m C_p) = T_f -T_i`

`T_i = T_f - (Q)/(m C_p)`

And replacing we got:

`T_f = 38.1C -(30 J)/(0.0131 Kg * 150 (J)/(Kg C))= 22.833 C`