Question

Question #bce5b

Answer 1

The specific heat capacity of the material is `"0.6469 J·°C"^"-1""g"^"-1"`.

Explanation:

The formula for the amount heat `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 = "-86 285 J"`
`m = "863.3 g"`
`ΔT = "(25.00 - 179.5) °C" = "-154.5 °C"`

∴ `C = "-86 285 J"/"863.3 g × (-154.5 °C)" = "0.6469 J·°C"^"-1""g"^"-1"`

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