Solid iron a specific heat of 0.444 J/g°C. If 75 calories of energy were removed from a piece of iron, and the temperature of the iron went from 21.4°C to -25.2°C, what was the mass of the iron?

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Solid iron a specific heat of 0.444 J/g°C. If 75 calories of energy were removed from a piece of iron, and the temperature of the iron went from 21.4°C to -25.2°C, what was the mass of the iron?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

When you look at the big picture, what you have is heat flowing out of solid iron and cooling it down.

So, you have to use the heat flow equation:

$\mathbf(q = msDeltaT)$

where:

$q$ is the heat flow in $"J"$.

$m$ is the mass of the object in $"g"$.

$s$ is the capacity of the object in $"J/g"cdot""^@ "C"$.

$DeltaT$ is the change in temperature of the iron in $""^@ "C"$.

First, convert the heat from $"cal"$ to $"J"$. Note that heat flowing out is negatively-signed by convention.

$-75$ $cancel"cal" xx "4.184 J"/cancel"cal" = -"313.8 J"$

Now what you have is:

$color(blue)(m_"Fe") = q_"Fe"/(s_"Fe"DeltaT_"Fe")$

$= q_"Fe"/(s_"Fe"[T_f^"Fe" - T_i^"Fe"])$

$= (-313.8 cancel("J"))/((0.444 cancel("J")"/g"cdotcancel(""^@ "C"))[(-25.2cancel(""^@ "C")) - (21.4cancel(""^@ "C"))])$

$=$ $color(blue)("15.2 g Fe")$

Solid iron a specific heat of 0.444 J/g°C. If 75 calories of energy were removed from a piece of iron, and the temperature of the iron went from 21.4°C to -25.2°C, what was the mass of the iron?

1 Answers