# 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

27.1 kJ

#### Explanation:

Very cold, no phase change. So, the heat change is just the amount of heat absorbed by the ice to raise its temperature by 29 degrees. The specific heat of ice is 2.100 J/g-’C.

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

The enthalpy change is thus 445g * (29’C) * 2.10 J/g-’C = 271000 Joules, (27.1 kJ).