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The specific heat capacity of zinc is 0.386 J/g ºC . How many joules would be released when 267 grams of zinc at 96.0 ºC were cooled to 28.0 ºC?

*Answer with a whole number*

Q = mcΔt

1 Answer

2 votes

Answer:

70008.216 J

Step-by-step explanation:

The following data were obtained from the question:

Specific heat capacity (C) = 0.386 J/gºC

Mass (m) = 267 g

Initial temperature (T1) = 96.0ºC

Final temperature (T2) = 28.0ºC

Change in temperature (ΔT) = T1 – T2 = 96 – 28 = 68°C

Heat (Q) =..?

The heat released can be obtained as follow:

Q = MCΔT

Q = 267 x 0.386 x 68

Q = 70008.216 J

Therefore, the heat released is 70008.216 J

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