When 24500J of energy is supplied to a 2.0 kg block of aluminum. Its temperature rise from 20oc to 35oc. The block is well insulate so that there is no energy loss to the sorrounding Determine specific aluminum

Explanation:

Amount of heat supplied (Q) = 24500 J

Mass (m) = 2 kg

Initial temperature (t1) = 20 ° C

Final temperature (t2) = 35 ° C

Difference in temperature (dt) = t2 - t1 = ( 35 - 20) °C = 15 ° C

Now Specific heat capacity (S) = ?

We know we have the formula;

Q = m * s * dt

24500 = 2 * s * 15

24500 = 30 S

S = 24500 / 30

S = 816.67 J / Kg ° C

The specific heat capacity is 816.67 J/kg°C.

Hope it will help :)

d is the correct answer

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