Question

5. A 5.00 g sample of an unknown substance was heated from 25.2 C to 55.1 degrees * C , and it required 133 to do so. Identify the unknown substance using the chart above.

show work please

Answer 1

Aluminum

Step-by-step explanation:

Given

`T_1 = 25.2^oC`

`T_2 = 55.1^oC`

`m = 5.00g`

`\triangle Q= 133J`

See attachment for chart

Required

Identify the unknown substance

To do this, we simply calculate the specific heat capacity from the given parameters using:

`c = (\triangle Q)/(m\triangle T)`

This gives:

`c = (\triangle Q)/(m(T_2 - T_1))`

So, we have:

`c = (133J)/(5.00g * (55.1C - 25.2C))`

`c = (133J)/(5.00g * 29.9C)`

`c = (133J)/(149.5gC)`

`c = 0.89\ J/gC`

From the attached chart, we have:

`Al(s) = 0.89\ J/gC` --- The specific heat capacity of Aluminum

Hence, the unknown substance is Aluminum