# A 445 g sample of ice at -58°C is heated until its temperature reaches -29°C. What is the change in heat content of the system?

May 20, 2017

The change in heat content is $\text{26,000 J}$.

#### Explanation:

Use the following equation:

$Q = m c \Delta T$,

where $Q$ is heat energy in Joules, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is change in temperature, $\left({T}_{\text{final"-T_"initial}}\right)$.

$\text{Organize your data}$.

Given/Known

$m = \text{445 g}$

c_"ice"=("2.03 J")/("g"*""^@"C")

${T}_{\text{initial"=-58^@"C}}$

${T}_{\text{final"=-29^@"C}}$

$\Delta T = - {29}^{\circ} \text{C"-(-58^@"C")="29"^@"C}$

Solution
Insert your data into the equation and solve.

Q=445color(red)cancel(color(black)("g"))xx(2.03"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C")))xx29^@color(red)cancel(color(black)("C"))="26,000 J" (rounded to two significant figures)

May 20, 2017

The temperature change is $- 29 - \left(- 58\right) = 29$