A 4.0 g sample of iron was heated from 0°C to 20°C. It absorbed 35.2 J of energy as heat. What is the specific heat of this piece of iron?

Md Ariful Islam

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Md Ariful Islam

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The formula for the heat absorbed by a substance is

$color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))" "$

where

$q$ is the quantity of heat

$m$ is the mass of the substance

$c$ is the specific heat capacity of the material

$ΔT$ is the temperature change

You can rearrange the formula to calculate the specific heat capacity:

$c = q/(mΔT)$

$ΔT = "(20 - 0) °C" = "20 °C"$

$c = "35.2 J"/"4.0 g × 20 °C" = "0.44 J·°C"^"-1""g"^"-1"$

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