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emmainna [20.7K]
3 years ago
13

385 J of heat are needed to heat a piece of aluminum from 22°C to 45°C. If the specific heat of Al is 0.90 J/g°C, what is the ma

ss of the aluminum sample?
Pls show work :))
Chemistry
1 answer:
VikaD [51]3 years ago
4 0

Answer:

M of Al=33.09g or 0.0331kg

Explanation:

Heat Energy= specific heat*mass*change in temperature

H=M*C*T

make M subject of the formula

M=H/CT

M=685J/0.90J/g°C*(45°C-22°C)

M=685J/0.90J/g°C*23°C

M=685J/20.7J/g

M=33.09g or 0.0331kg

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Answer:

We have to take 37.5 mL of a 0.400 M solution

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