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A 126-gram sample of titanium metal is heated from 20.0°C to 45.4°C while absorbing 1.68 kJ of heat. What is the specific heat of titanium? a 0.667 J/g⋅∘C b 0.294 J/g⋅∘C c 0.525 J/g⋅∘C d 0.471 J/g⋅∘C

Question

A 126-gram sample of titanium metal is heated from 20.0°C to 45.4°C while absorbing 1.68 kJ of heat. What is the specific heat of titanium? a 0.667 J/g⋅∘C b 0.294 J/g⋅∘C c 0.525 J/g⋅∘C d 0.471 J/g⋅∘C

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Solution

The specific heat of a substance can be calculated using the formula:

q = mcΔT

where: q = heat absorbed or released (in joules or J) m = mass of the substance (in grams or g) c = specific heat (in J/g°C) ΔT = change in temperature (in °C)

In this case, we are given: q = 1.68 kJ = 1680 J (since 1 kJ = 1000 J) m = 126 g ΔT = 45.4°C - 20.0°C = 25.4°C

We are asked to find c, so we rearrange the formula to solve for c:

c = q / (mΔT)

Substituting the given values:

c = 1680 J / (126 g * 25.4°C) = 0.525 J/g°C

So, the specific heat of titanium is 0.525 J/g°C, which corresponds to option c.

This problem has been solved

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