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)

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Answer 1 · 2018-03-02T03:54:35

Total Energy required is: $5727.315 J$ $5727.315J$

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

Formula: $E = m L$ $E=mL$

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

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

Formula: $E = m c T$ $E=mcT$

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

$E = m c T$ $E=mcT$

$E = 10 g \cdot 5^{o} C \cdot 2.108 \frac{J}{g^{o} C}$ $E=10g * 5^oC * 2.108J/(g^oC)$

$E = 105.4 J$ $E=105.4J$

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

$E = m L$ $E=mL$

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

$E = 3340 J$ $E=3340J$

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

$E = m c T$ $E=mcT$

$E = 10 g \cdot {54.5}^{o} C \cdot 4.187 \frac{J}{g^{o} C}$ $E=10g * 54.5^oC * 4.187J/(g^oC)$

$E = 2281.915 J$ $E=2281.915J$

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

$105.4 J + 3340 J + 2281.915 J$ $105.4J+3340J+2281.915J$

$= 5727.315 J$ $=5727.315J$

And there we have our answer.