Question

How many kilojoules of energy are needed to convert 89.0 g of ice at -19.5 to water at 26.8°C? (The specific heat of ice at -19.5 is 2.01 J/g°C.)

Answer 1

Answer:

8.28kJ

Explanation:

We'll use our formula for specific heat Q=mcΔT.

Q= heat (Joules), m= mass (grams),

ΔT= change in temperature (Final temp.- initial temp.)

1. We're solving for Q because we're looking for energy in the form of heat.

Let's plug in what we know.

Q=(89g)(2.01 J/g C)(26.8C- (-19.5C))

Q=(89g)(2.01 J/g C)(26.8 + 19.5)

Q=(89g)(2.01 J/g C)(46.3 C)

Q=8283 J

2. Let's check our units first before submitting our answer, they're asking for it to be in kilojoules so we'll convert J -> kJ.

1000 Joules = 1 kJ

8283J x $\frac{1 kJ}{1000J}$ = 8.28 kJ