How many joules of heat are removed to completely covert 80.0 grams of water to ice at zero Celcius?
For water: c = 4.179 J/g °C Hfusion = 333.528J/g Hvap = 2258.601 J/g
q = mass H2O x heat fusion
To find the amount of heat removed to completely convert 80.0 grams of water to ice at zero degrees Celsius, we need to consider two processes: cooling the water from its initial temperature to zero degrees Celsius and then freezing the water at zero degrees Celsius.
Let's break down the calculation step-by-step:
1. Calculate the heat removed to cool the water from its initial temperature to zero degrees Celsius.
- To do this, we will use the specific heat capacity of water (c) and the formula: q = m * c * ΔT, where q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- In this case, the initial temperature is not specified, so we will assume it is 25 degrees Celsius.
- ΔT is then (0 - 25) = -25 degrees Celsius.
Calculating the heat removed for cooling:
q1 = m * c * ΔT
q1 = 80.0 g * 4.179 J/g °C * (-25 °C)
q1 = -83,580 J
2. Calculate the heat removed for freezing the water at zero degrees Celsius.
- This step involves the heat of fusion (Hfusion), which is the energy required to convert one gram of a substance from liquid to solid at its melting point.
- The formula is: q = m * Hfusion, where q is the heat energy and m is the mass.
Calculating the heat removed for freezing:
q2 = m * Hfusion
q2 = 80.0 g * 333.528 J/g
q2 = 26,682.24 J
3. Total heat removed:
The total heat removed is the sum of the heat removed for cooling (q1) and the heat removed for freezing (q2).
q total = q1 + q2
q total = -83,580 J + 26,682.24 J
q total = -56,897.76 J
Therefore, approximately 56,898 joules of heat are removed to completely convert 80.0 grams of water to ice at zero degrees Celsius.