# Imagine the 10.0g of ice has an initial temperature of -5.0degreesC. Can someone calculate the energy needed to heat the ice, melt it, and heat the water to 54.5degreesC? (Cp of ice=2.108J/gdegreesC)

Mar 2, 2018

Total Energy required is: $5727.315 J$

#### Explanation:

Latent Heat of Fusion of Ice: $334 \frac{J}{g}$ ($L$)

Formula: $E = m L$

Specific Heat Capacity of Ice: $2.108 \frac{J}{{g}^{o} C}$ ($c$)

Specific Heat Capacity of Water: $4.187 \frac{J}{{g}^{o} C}$ ($c$)

Formula: $E = m c T$

To calculate this, first we must find the energy required to raise the temperature of ice from $- {5}^{o} C$ to it's melting point:

$E = m c T$

$E = 10 g \cdot {5}^{o} C \cdot 2.108 \frac{J}{{g}^{o} C}$

$E = 105.4 J$

Now, we must find the Energy required to convert the solid ice into liquid water at its melting point:

$E = m L$

$E = 10 g \cdot 334 \frac{J}{g}$

$E = 3340 J$

Finally, we must find the Energy required to raise the temperature of water from ${0}^{o} C$ to ${54.5}^{o} C$:

$E = m c T$

$E = 10 g \cdot {54.5}^{o} C \cdot 4.187 \frac{J}{{g}^{o} C}$

$E = 2281.915 J$

To calculate the total Energy needed, we need to add all the energy values up:

$105.4 J + 3340 J + 2281.915 J$

$= 5727.315 J$

And there we have our answer.