Calorimetry: Heat Lost by Cooling Coffee

Given

• Mass of coffee: m = 250.0 g

• Specific heat capacity: c = 4.184 J / (g degC) (coffee treated as water)

• Initial temperature: 90.0 degC; final temperature: 60.0 degC

Find

The heat change q in kJ, with the correct sign.

Principle

Calorimetry tracks heat through q = mc delta T: the heat gained or lost equals mass times specific heat capacity times the temperature change. The sign of delta T (final minus initial) gives the sign of q — negative when the object cools (heat leaves), positive when it warms (heat enters). Specific heat capacity is the material's personal rate card for heat per gram per degree.

Formula

Solution

1. Temperature change: delta T = T_final - T_initial = 60.0 - 90.0 = -30.0 degC.

2. Apply q = m c delta T: q = 250.0 g x 4.184 J/(g degC) x (-30.0 degC).

3. Multiply c x delta T first: 4.184 x (-30.0) = -125.52 J/g.

4. Multiply by mass: 250.0 x (-125.52) = -31,380 J.

5. Convert to kilojoules: -31,380 J / 1000 = -31.4 kJ.

6. The negative sign means heat left the coffee: it released 31.4 kJ.

Unit check

g x (J/(g degC)) x degC leaves joules — grams and degrees cancel exactly. Converting J to kJ is a factor-of-1000 unit shift with no chemistry changed.

What this means physically

Letting a cup cool from 90 to 60 degC dumps about 31 kJ into the room — roughly the energy of a small snack. This is also why drinks cool fastest when the temperature gap is largest: heat flow is proportional to delta T.