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The specific heat of iron is 0.473 J/gºC. How much energy is required to heat a 40.0 g sample of iron from 35.0 ºC to 75.0 ºC?

a) 757 J
b) 2080 J
c) 6690 J
d) 1320 J

User Violet
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1 Answer

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Final answer:

The energy required to heat a 40.0 g sample of iron from 35.0 °C to 75.0 °C is approximately 757 J, according to the formula q = mcΔT with the specific heat of iron being 0.473 J/g°C. Option A is correct.

Step-by-step explanation:

The question involves calculating the energy required to heat a sample of iron given its mass, temperature change, and specific heat. To find the solution, we can apply the formula q = mcΔT, where:

q is the heat energy (in joules)

m is the mass of the substance (in grams)

c is the specific heat capacity (J/g°C)

ΔT is the change in temperature (°C)

The specific heat of iron is 0.473 J/g°C, the mass of the sample is 40.0 g, and the temperature change (ΔT) is from 35.0 °C to 75.0 °C, which is a change of 40.0 °C.

By substituting these values into the formula:

q = (40.0 g) ∗ (0.473 J/g°C) ∗ (40.0 °C)

We get: q = 40.0 g ∗ 0.473 J/g°C ∗ 40.0 °C = 753.6 J, which rounds to 757 J when considering significant figures.

So, the energy required to heat the sample of iron from 35.0 °C to 75.0 °C is approximately 757 J, which would correspond to option (a).

User Yogesh Pingle
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