Need ans asap plsss if you supply 3600 kj of heat, how many grams of ice at 0°c can be melted, heated to its boiling point? (make m your x in equations) (figure out which q you need)

17,140 g = 17.14 kg.

Explanation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).

m is the mass of the ice (m = ??? g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).

∵ Q = m.c.ΔT

∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)

∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.

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