A piece of gold initially at a temperature of 25.1 °C absorbs 675 J of heat, raising its temperature to 57.4 °C. Assuming the specific heat of gold is 0.126 J/(g°C), what is the mass of the sample?
The key here is the specific heat of gold, which is said to be equal to
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Using the formula for heat transfer:
Q = mcΔT
where: Q = heat transferred (in joules) m = mass of the sample (in grams) c = specific heat capacity of gold (in J/(g°C)) ΔT = change in temperature (in °C)
Rearranging the formula to solve for mass:
m = Q / (c * ΔT)
Plugging in the given values:
m = 675 J / (0.126 J/(g°C) * (57.4°C - 25.1°C))
m ≈ 25.96 grams
Therefore, the mass of the sample is approximately 25.96 grams.
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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.
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.
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.
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.
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