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How much heat, in calories, is given off when 1.25 grams of silver is cooled from 100.0 C to 80.0 C? (The specific heat of silver is 0.057 cal/g C)?

Answer 1

Luna Chen

The heat is #=1.425cal#

The heat is

#Q=m*s*DeltaT#

mass, #m=1.25g#

specific heat, #s=0.057calg^-1ºC^-1#

#Delta T=100-80=20ºC#

Heat liberated is #=1.25*0.057*20=1.425cal#

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Answer 2

Avery Adams

To calculate the heat given off when cooling 1.25 grams of silver from 100.0°C to 80.0°C, you can use the formula:

Q = mcΔT

Where: Q = heat transferred (in calories) m = mass of the substance (in grams) c = specific heat capacity of the substance (in cal/g°C) ΔT = change in temperature (in °C)

Substituting the given values:

m = 1.25 g c = 0.057 cal/g°C ΔT = (80.0°C - 100.0°C) = -20.0°C

Q = (1.25 g) * (0.057 cal/g°C) * (-20.0°C) Q = -14.25 cal

So, when 1.25 grams of silver is cooled from 100.0°C to 80.0°C, it gives off 14.25 calories of heat.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.