A 136 g sample of an unknown substance was heated from 20.0 °C to 40.0 °C. In the process the substance absorbed 2440 J of energy. What is the specific heat of the substance? Identify the substance among those listed in Table 2. A. the specific heat is 0.897 J/g. C, The Substance is aluminum
B. the specific heat is -0.897 J/g. C, The Substance is aluminum
C. the specific heat is 4.184 J/g. C, The Substance is water
D. there's not enough information to determine which is the substance.

Option A. The specific heat is 0.897 J/gºC, The Substance is aluminum.

Explanation:

We'll begin by calculating the change in temperature of the substance. This can be obtained as follow:

Initial temperature (T₁) = 20.0 °C

Final temperature (T₂) = 40.0 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 40 – 20

ΔT = 20 °C

Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:

Mass (M) = 136 g

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 2440 J

Specific heat capacity (C) =?

Q = MCΔT

2440 = 136 × C × 20

2440 = 2720 × C

Divide both side by 2720

C = 2440 / 2720

C = 0.897 J/gºC

Comparing the specific heat capacity (i.e 0.897 J/gºC) of the substance with those in the table above, the substance is Aluminum.

Thus, option A gives the correct answer to the question.

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