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How many joules would be released when 256 grams of zinc at 96 c were cooled to 28 c

1 Answer

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Answer:

–6562.816 J

Step-by-step explanation:

From the question given above, the following data were obtained:

Msss (M) = 256 g

Initial temperature (T₁) = 96 °C

Final temperature (T₂) = 28 °C

Heat (Q) released =?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial temperature (T₁) = 96 °C

Final temperature (T₂) = 28 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 28 – 96

ΔT = –68 °C

Finally, we shall determine the heat released. This can be obtained as follow:

Msss (M) = 256 g

Change in temperature (ΔT) = –68 °C

Specific heat capacity (C) of Zn = 0.377 J/gºC

Heat (Q) released =?

Q = MCΔT

Q = 256 × 0.377 × –68

Q = –6562.816 J

Thus, –6562.816 J of heat energy is released.

User Giovanni Di Milia
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