How much heat energy, in kilojoules, is required to convert 48.0g of ice at −18.0 ∘C to water at 25.0 ∘C ?

Question

Please show all steps. i don't know where to start

Answer 1

There are three equations that handle all of these situations.
Equation 1 is for WITHIN a phase,
q = mass x specific heat x (Tfinal-Tinitial).
Example. For the liquid phase of water from zero C to 25 C, then q for that part is
q = (48g x 4.184 J/g*C x (25-0) = ?

Equations 2 and 3 are at a phase change (for water that is melting ice at zero C or boiling at 100 C to steam).
q = mass x heat fusion for melting or
q = mass x heat vaporization for boiling.

Example. For melting ice at zero to liquid water (still at zero)
q = 48g ice x 334 J/g.

So you have to move ice at -18.0 C to 25.0 C and you do it in steps.

q1 = heat to raise T of ice at -18.0C to ice at zero C. That's equation 1.

q2 = heat to melt ice at zero to liquid at zero. That's equation 2.

q3 = heat to raise T of liquid water from zero C to 25 C. That's equation 1.

Total q is q1 + q2 + q3.

Answer 2

21.4