How much heat energy is released to surrounding when 15.0 g of water at 35°C is converted into ice with a temperature of -45°C?

vap = 2260 J/g; Hfus = 334 J/g; Cp(ice) = 2.06 J/g • °C; Cp(water) = 4.18 J/g • °C;
Cp(steam) = 1.99 J/g • °C

Similar to your first post. You work it out. Post your work if you get stuck.

46554

To find the amount of heat energy released when water is converted into ice, we need to consider the following steps:

1. Calculate the heat energy released to cool the water from 35°C to 0°C.
2. Calculate the heat energy released during the phase change from water at 0°C to ice at 0°C.
3. Calculate the heat energy released to further cool the ice from 0°C to -45°C.

Let's break down these steps and calculate each part:

Step 1: Calculate the heat energy released to cool the water from 35°C to 0°C.
We'll use the formula: Q = m * Cp * ΔT
where Q is the heat energy, m is the mass, Cp is the specific heat capacity, and ΔT is the change in temperature.

Given:
Mass of water (m) = 15.0 g
Initial temperature (T1) of water = 35°C
Final temperature (T2) of water = 0°C
Specific heat capacity (Cp) of water = 4.18 J/g • °C

Using the formula Q = m * Cp * ΔT, we can calculate the heat energy released during this step:
Q1 = 15.0 g * 4.18 J/g • °C * (0°C - 35°C)
Q1 = -2208.75 J

So, the heat energy released to cool the water from 35°C to 0°C is -2208.75 J. The negative sign indicates energy release.

Step 2: Calculate the heat energy released during the phase change from water at 0°C to ice at 0°C.
We'll use the formula: Q = m * ΔHfus
where Q is the heat energy, m is the mass, and ΔHfus is the heat of fusion.

Given:
Mass of water (m) = 15.0 g
Heat of fusion (ΔHfus) = 334 J/g

Using the formula Q = m * ΔHfus, we can calculate the heat energy released during this step:
Q2 = 15.0 g * 334 J/g
Q2 = 5010 J

So, the heat energy released during the phase change from water at 0°C to ice at 0°C is 5010 J.

Step 3: Calculate the heat energy released to further cool the ice from 0°C to -45°C.
We'll use the formula: Q = m * Cp * ΔT
where Q is the heat energy, m is the mass, Cp is the specific heat capacity, and ΔT is the change in temperature.

Given:
Mass of ice (m) = 15.0 g
Initial temperature (T1) of ice = 0°C
Final temperature (T2) of ice = -45°C
Specific heat capacity (Cp) of ice = 2.06 J/g • °C

Using the formula Q = m * Cp * ΔT, we can calculate the heat energy released during this step:
Q3 = 15.0 g * 2.06 J/g • °C * (-45°C - 0°C)
Q3 = -1543.5 J

So, the heat energy released to further cool the ice from 0°C to -45°C is -1543.5 J.

Now, let's sum up the heat energy released during each step to find the total heat energy released:

Total heat energy released = Q1 + Q2 + Q3
Total heat energy released = -2208.75 J + 5010 J - 1543.5 J
Total heat energy released = 1257.75 J

Therefore, the total heat energy released to the surroundings when 15.0 g of water at 35°C is converted into ice with a temperature of -45°C is approximately 1257.75 J.