Question

How much energy (in Joules) is required to convert 129 grams of ice at −23.0 °C to liquid water at 18.0 °C?

(The specific heat of ice = 2.10 J/g°C; the specific heat of water = 4.184 J/g°C; the heat of fusion = 333 J/g°C).

Answer 1

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules

Step-by-step explanation:

This is a calorimetry problem:

Q = m . C . ΔT

Q = heat; m = mas; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

First of all we calculate the heat for ice, before it takes the melting point. (from -23°C to 0°C)

Q = 129 g . 2.10 J/g°C . (0°C - (-23°C)

Q = 129 g . 2.10 J/g°C . 23°C → 6230.7 joules

Then, the ice has melted. To be melted and change the state it required:

Q = C lat . m

Q = 333 J/°C . 129 g → 42957 joules

And in the end, we have water that changed its T° from O°C to 18°C

Q = 129 g . 4.184 J/g °C . (18°C - 0°C)

Q = 9715.2 Joules

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules