# How much energy would be absorbed as heat by 75g of iron when heated from 295k to 301k? c=.46

Jul 24, 2018

$\text{210 Joules}$ of heat energy are required.

#### Explanation:

Equation:

$q = m c \Delta T$,

where:

$q$ is energy, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is change in temperature. $\Delta T = {T}_{\text{final"-T_"initial}}$

Known

$c = \text{0.46 J/g"*"K}$

$m = \text{75 g}$

$\Delta T = \text{301 K"-"295 K"="6 K}$

Unknown

energy, $q$

Plug the known values into the equation and solve.

q=75color(red)cancel(color(black)("g"))xx0.46("J")/(color(red)cancel(color(black)("g"))*color(red)cancel(color(black)("K")))xx6color(red)cancel(color(black)("K"))="210 J" (rounded to two significant figures)

$\text{210 Joules}$ of heat energy are required.