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Given 28.2 grams of an unknown substance, if the substance absorbs 2165 joules of energy and the temperature increases by 35 Kelvin, what is the specific heat of the substance? 2.14 x 106 J/g·K 2.19 x 100 J/g·K 5.73 x 10-4 J/g·K 2.69 x 103 J/g·K If someone could tell me how to do it as well, it would be greatly appreciated.

Respuesta :

lulu1991
I think the answer should be 2.19 J/g*K. The specific heat means the energy absorbed by 1 g substance to increase 1 Kelvin temperature. And keep the unit the same.

Answer:

The specific heat of the substance is 2.19 J/gK

Explanation:

Given:

Mass of the substance, m = 28.2 g

Heat absorbed, q = 2165 J

The change in temperature, ΔT = 35 K

To determine:

The specific heat c of the substance

Explanation:

The amount of heat (q) absorbed (or evolved) by  a substance of mass (m) in order to bring a temperature change of (ΔT) is given as:

[tex]q = mc\Delta T[/tex]

[tex]c = \frac{q}{m\Delta T}[/tex]

In this case:

c = \frac{2165 J}{28.2g *35K }=2.19 J/g K

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