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A 2.15 kg block of aluminum (specific heat = 897 J/kg·K) is at an initial temperature of 300 K. What will its final temperature be if 335,000 J of thermal energy are added?

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samueladesida43

The final temperature of the aluminum block is 473.71K

SPECIFIC HEAT CAPACITY:

  • The quantity of heat absorbed or released by a substance can be calculated using the following formula:

  • Q = m × c × ∆T

Where;

Q = quantity of heat absorbed/released (J)

m = mass of the substance (kg)

c = specific heat (J/kgK)

∆T = change in temperature (K)

  • The information given in this question is as follows:

Q = 335,000J

m = 2.15kg

c = 897 J/kg·K

∆T = T2 - T1

  • Q = m × c × ∆T

335,000 = 2.15 × 897 × (T2 - 300)

335,000 = 1928.55 (T2 - 300)

335,000 = 1928.55T2 - 578565

335000 + 578565 = 1928.55T2

913565 = 1928.55T2

T2 = 913565 ÷ 1928.55

T2 = 473.71K

Therefore, the final temperature of the aluminum block is 473.71K.

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