How much heat is released when 28.95g of water is frozen?

q = mass x heat fusion

34j

To calculate the amount of heat released when water is frozen, we need to use the specific heat capacity and the heat of fusion of water.

The specific heat capacity of water is 4.18 J/g·°C, which means it takes 4.18 Joules of heat per gram of water to change the temperature by 1 degree Celsius.

The heat of fusion of water is 333.5 J/g, which represents the amount of heat required to freeze 1 gram of water into ice at 0 degrees Celsius without changing its temperature.

Since we want to find the heat released when 28.95 grams of water is frozen, we need to calculate the heat required to change the water from its initial temperature (let's assume it is 25 degrees Celsius) to the freezing point (0 degrees Celsius), and then calculate the heat released during the freezing process.

First, we calculate the heat required to lower the temperature of 28.95 grams of water from 25 degrees Celsius to 0 degrees Celsius:

Heat = mass × specific heat capacity × change in temperature
Heat = 28.95 g × 4.18 J/g·°C × (0 - 25) °C

Next, we calculate the heat released during the freezing process:

Heat = mass × heat of fusion
Heat = 28.95 g × 333.5 J/g

Finally, we add these two amounts of heat together to find the total heat released:

Total heat released = heat required to lower temperature + heat released during freezing

Note: Please provide the initial temperature of the water if it's different from 25°C, and we can calculate the exact amount of heat released.

To determine the amount of heat released when water freezes, we need to use the concept of heat transfer. The formula we can use is:

q = m * ΔH

Where:
q is the heat released or absorbed (in Joules)
m is the mass of the substance (in grams)
ΔH is the enthalpy change per gram (in J/g)

In this case, we need to know the enthalpy change of water when it freezes. The enthalpy change for water freezing is typically given as 334 J/g, so we will use that value.

Using the formula, we can calculate the heat released when 28.95 g of water freezes:

q = 28.95 g * 334 J/g
q = 9672.3 J

Therefore, when 28.95 g of water freezes, approximately 9672.3 Joules of heat are released.