# How much energy is required to raise the temperature of a "36.00 g" sample of ice from a temperature of -30.00^@"C" to a temperature of "85.00"^@"C"? The heat of fusion, H_"f", of water is "334 J/g".

Jul 28, 2014

The total energy required is $\text{15577 J}$ or $\text{15.577 kJ}$.

#### Explanation:

The basic equation used to determine the amount of energy needed to change the temperature of a given mass of a substance is:

$q = m \times c \times \Delta T$,

where:

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

There are four steps required to answer this question. First, the energy required to raise the temperature of the ice from $- {30.00}^{\circ} \text{C}$ to ${0.00}^{\circ} \text{C}$ must be determined. Then the energy required to melt the ice must be determined. Then the energy required to raise the temperature from ${0.00}^{\circ} \text{C}$ to ${85.00}^{\circ} \text{C}$ must be determined. Once you have all three values for energy, add them together to get the total energy.

1. Energy required to change the temperature from $- {30.00}^{\circ} \text{C}$ to ${0}^{\circ} \text{K}$. In this equation we will use the specific heat of solid water (ice).

Known

$m = \text{36.00 g}$

c_"ice"=(2.108"J")/("g"*""^@"C")

${T}_{\text{i"=-30.00^@"C}}$

${T}_{\text{f"=0.00^@"C}}$

$\Delta T = {0.00}^{\circ} \text{C"-(- 30.00^@"C")="30.00"^@"C}$

Solution

Plug the known values into the equation and solve for $q$.

q = (36.00color(red)cancel(color(black)("g")))xx((2.108"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C"))))xx(30.00^@color(red)cancel(color(black)("C")))="2277 J" (rounded to four significant figures)

2. Energy required to melt the ice.

Equation:

$q = \text{m"xxH_"f}$,

where:

$q$ is energy, ${H}_{\text{f}}$ is the heat of fusion, and $m$ is mass.

The temperature is not needed because the temperature doesn't change during a phase change.

Known

${H}_{\text{f"="334 J/g}}$

$m = \text{36.00 g}$

Solution

Plug in the known values and solve for $q$.

q=36.00color(red)cancel(color(black)("g"))xx(334 "J")/color(red)cancel(color(black)("g"))="12020 J" (rounded to four significant figures)

3. Energy required to change the temperature from $\text{0.00"^@"C}$ to $\text{85"^@"C}$. In this equation we will use the specific heat for liquid water.

Equation:

$q = m c \Delta T$

Known

${c}_{\text{H2O"=(4.184 "J")/("g"*""^@"C")}}$

$m = \text{36.00 g}$

${T}_{\text{i"="0.00"^@"C}}$

${T}_{\text{f"="85.00"^@"C}}$

$\Delta T = \text{85.00"^@"C"-"0.00"^@"C"="85.00"^@"C}$

Solution

Plug in the known values and solve for $q$.

q=36.00color(red)cancel(color(black)("g"))xx(4.184"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C")))xx85.00^@color(red)cancel(color(black)("C"))= "1280. J" (rounded to four significant figures)

4. Add the energies from each of the three equations. This will give you the total amount of energy needed to raise the temperature of a $\text{36.00 g}$ sample of ice from $- {30.00}^{\circ} \text{C}$ to $\text{85.00"^@"C}$.

$\text{Total energy"="2277 J" + "12020 J" + "1280 J"="15577 J}$ (rounded to a whole number)