Melting ice using latent heat and heat of fusion

[Navygal]

Homework Statement

What quantity of ice at 2 deg Celsius can be melted by 100 J of heat?

Homework Equations

q = m c (delta) t (mass x specific heat x change in temp)
q = m l (mass x heat of fusion)

specific heat of ice = 2.09 j/g-K
heat of fusion = 6.01 kJ/mol

The Attempt at a Solution

I've tried to solve this problem by combining both equations for the same q such that
q = mc(delta)t + ml which rearranges to:
q = m (c delta t + l)

solving for m: m = q/ (c delta t + l)

The part I got stuck is when I tried to solve for m since m is in grams, and the heat of fusion is in moles. Is there another way to solve this or am I missing something?
Any help would be greatly appreciated

 

[Ygggdrasil]

You can use the molecular weight of water (~18g/mol) to convert between grams and moles.

 

[Blueshift5]

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I would first clarify that the heat of fusion is the amount of heat required to melt a substance at its melting point without changing its temperature. In this case, we are dealing with ice at 2 degrees Celsius, which is already below its melting point. Therefore, we can use the specific heat equation to calculate the amount of heat needed to raise the temperature of the ice to its melting point (0 degrees Celsius). This would be equal to q = m c (delta) t, where q = 100 J, c = 2.09 J/g-K, (delta) t = (0-2) = -2 K.

Next, we can use the heat of fusion equation to calculate the amount of heat needed to melt the ice at its melting point. This would be equal to q = m l, where q = 100 J, l = 6.01 kJ/mol. However, we need to convert the units of l from kJ/mol to J/g by dividing by the molar mass of ice (18.02 g/mol). This gives us l = 333.5 J/g.

Now, we can combine the two equations to solve for the mass of ice (m) that can be melted by 100 J of heat. This would be equal to m = q/ (c delta t + l), which gives us m = 100 J / (2.09 J/g-K x -2 K + 333.5 J/g) = 0.301 g.

Therefore, 100 J of heat can melt 0.301 g of ice at 2 degrees Celsius.