How many joules of heat are needed to heat 25.0 g of ice at -5.0 degrees celcius to a final temperature of 40.0 degrees celcius? The specific heat of ice is 2.1 j. Our teacher mention to consider the state of water present at 40.0 degrees celcius which is confusing me.

Question

Any help is appreciated. Thanks.

Answer 1

You go through each state. At -5 C it is solid and goes to zero as ice.
q1 = heat absorbed in moving ice from -5C to zero.
q1 = mass ice x specific heat ice x (Tfijal-Tinitial)

q2 = heat absorbed in melting ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion.

q3 = heat absorbed in moving liquid water from zero C to 40 C.
q3 = mass water x specific heat water x (Tfinal-Tintial)
Total Q = q1 + q2 + q3.

You see this is just two things you need to worry about.
When moving WITHIN a phase you use
q = mass x specific heat x (Tfinal-Tintial)
When changing phase at a point (a certain T), one uses q = mass x delta H where delta H is Heat fusion for melting or heat vaporization for boiling. So one can go from ice at -50 C to steam at 150C and it sounds so confusing BUT it needs just those two formulas to do it as long as you remember what phase you're in and where the phase changes are, then add the individual q values together to obtain final Q.