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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

Sagot :

MrRoyal

Answer:

Aluminum

Explanation:

Given

[tex]T_1 = 25.2^oC[/tex]

[tex]T_2 = 55.1^oC[/tex]

[tex]m = 5.00g[/tex]

[tex]\triangle Q= 133J[/tex]

See attachment for chart

Required

Identify the unknown substance

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

[tex]c = \frac{\triangle Q}{m\triangle T}[/tex]

This gives:

[tex]c = \frac{\triangle Q}{m(T_2 - T_1)}[/tex]

So, we have:

[tex]c = \frac{133J}{5.00g * (55.1C - 25.2C)}[/tex]

[tex]c = \frac{133J}{5.00g * 29.9C}[/tex]

[tex]c = \frac{133J}{149.5gC}[/tex]

[tex]c = 0.89\ J/gC[/tex]

From the attached chart, we have:

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

Hence, the unknown substance is Aluminum

View image MrRoyal
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