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

Answer:

Aluminum

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:

$c = \frac{\triangle Q}{m\triangle T}$

This gives:

$c = \frac{\triangle Q}{m(T_2 - T_1)}$

So, we have:

$c = \frac{133J}{5.00g * (55.1C - 25.2C)}$

$c = \frac{133J}{5.00g * 29.9C}$

$c = \frac{133J}{149.5gC}$

$c = 0.89\ J/gC$

From the attached chart, we have:

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

Hence, the unknown substance is Aluminum