Question #b5be7

Question

2047 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-19T19:07:36

The specific heat capacity of the metal is ${0.11 cal\cdot°C}^{-1} g^{-1}$ $"0.11 cal·°C"^"-1""g"^"-1"$ .

The formula for the heat $q$ $q$ absorbed by a substance is

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

where

$m$ 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:

$c = q/(mΔT)$

In this problem,

$q = "26.5 cal"$
$m = "50 g"$
$ΔT = "5 °C"$

∴ $c = "26.5 cal"/"50 g × 5 °C" = "0.11 cal·°C"^"-1""g"^"-1"$