For a 8.88 gram sample of ice at -43.6°C , what is the minimum amount of heat (in kJ) needed to convert the sample to liquid water at 0.00°C?

Question

For a 8.88 gram sample of ice at -43.6°C , what is the minimum amount of heat (in kJ) needed to convert the sample to liquid water at 0.00°C?

Physical data for water:
specific heat (ice) = 2.10 J/g°C
specific heat (water) = 4.184 J/g°C
heat of fusion = 0.335 kJ/g
heat of vaporization = 2.258 kJ/g

Answer 1

q1 = heat to warm ice from -43.6 C to zero C.

q1 = mass ice x specific heat ice x (Tfinal - Tinital) = ?

q2 = heat to melt the ice
q2 = mass ice x heat fusion ice = ?

q total = q1 + q2 = ?. Note that q1 answer is in J and q2 is in kJ. Since the problem asks for the answer in kJ, you should calculatae q1 and convert to kJ before adding to q2. Post your work if you get stuck.

Answer 2

1) If the specific heat capacity of ice is 0.5 cal/gC°, how much heat would have to be added to 200 g of ice, initially at a temperature of -10°C, to raise the ice to the melting point?

Answer 3

Well, well, well, it looks like we have some "cool" chemistry here! Let's break it down, shall we?

To convert the ice sample to liquid water, we need to go through three steps: heating the ice to 0°C, melting the ice into water, and then heating the water to 0°C.

First, let's heat the ice from -43.6°C to 0°C. We can use the formula:

Q = m * C * ΔT

where Q is the heat energy, m is the mass, C is the specific heat, and ΔT is the change in temperature.

For this step, we have:
m = 8.88 g
C = 2.10 J/g°C
ΔT = 0°C - (-43.6°C) = 43.6°C

Plugging these values into the formula, we get:
Q1 = 8.88 g * 2.10 J/g°C * 43.6°C = 822.14 J

Now, let's melt the ice into water. The heat of fusion tells us how much heat is needed to convert 1 gram of ice into water. We multiply this by the mass of the ice to find the total heat required to melt it. In this case, the heat of fusion is 0.335 kJ/g, so:

Q2 = 8.88 g * 0.335 kJ/g = 2.9772 kJ

Lastly, let's heat the water from 0°C to 0°C (yes, I know, it sounds weird). Using the same formula as before:

Q3 = 8.88 g * 4.184 J/g°C * 0°C = 0 J (Heating water with the same final temperature means no temperature change)

Now, let's sum up the heat required for each step to find the total minimum amount of heat:

Total heat = Q1 + Q2 + Q3
Total heat = 822.14 J + 2.9772 kJ + 0 J

Drumroll, please...

Total heat = 3.799 kJ

And voilà! We have our answer - the minimum amount of heat needed to convert the ice sample to liquid water at 0.00°C is approximately 3.799 kJ. Stay cool!

Answer 4

To find the minimum amount of heat required to convert the sample from ice at -43.6°C to liquid water at 0.00°C, we need to consider three steps:

Step 1: Heating the ice from -43.6°C to 0.00°C
Step 2: Melting the ice at 0.00°C
Step 3: Heating the liquid water from 0.00°C to its final temperature (if needed)

Let's calculate each step individually and then sum them up to find the total heat required.

Step 1: Heating the ice from -43.6°C to 0.00°C
To calculate the heat required in this step, we will use the formula: Q = m * c * ΔT, where:
Q = heat required
m = mass of the sample (8.88 g)
c = specific heat capacity of ice (2.10 J/g°C)
ΔT = change in temperature (0.00°C - (-43.6°C))

Using the formula, we can calculate the heat required in this step:
Q1 = 8.88 g * 2.10 J/g°C * (0.00°C - (-43.6°C))
Q1 = 8.88 g * 2.10 J/g°C * 43.6°C

Step 2: Melting the ice at 0.00°C
To calculate the heat required to melt the ice, we will use the formula: Q = m * ΔHf, where:
Q = heat required
m = mass of the sample (8.88 g)
ΔHf = heat of fusion for ice (0.335 kJ/g)

Using the formula, we can calculate the heat required in this step:
Q2 = 8.88 g * 0.335 kJ/g

Step 3: Heating the liquid water from 0.00°C to its final temperature (if needed)
Since the final temperature is not specified, we will assume it is the same as the initial temperature, 0.00°C. Therefore, no additional heat is required in this step.

Now, let's sum up the heat required in each step to find the total heat required:
Total heat required = Q1 + Q2

Please substitute the values into the calculations and solve to find the minimum amount of heat required to convert the sample from ice to water.

Answer 5

To solve this problem, we need to consider the following steps:

1. Calculate the heat required to warm the ice from -43.6°C to 0.00°C.
- We will use the formula: q = m * C * ΔT, where q is the heat, m is the mass, C is the specific heat, and ΔT is the change in temperature.
- Calculate the heat for this step using the specific heat of ice (2.10 J/g°C).
- Convert the result to kJ by dividing by 1000.

2. Calculate the heat required to melt the ice at 0.00°C.
- We will use the formula: q = m * ΔH, where q is the heat, m is the mass, and ΔH is the heat of fusion.
- Calculate the heat for this step using the heat of fusion (0.335 kJ/g).

3. Add the heat calculated in step 1 and step 2 to get the total heat required.

Let's calculate the heat required for each step:

Step 1:
q1 = m * C * ΔT
q1 = 8.88 g * 2.10 J/g°C * (0.00°C - (-43.6°C))
q1 = 8.88 g * 2.10 J/g°C * 43.6°C
q1 = 822.288 J
q1 = 822.288 J / 1000 = 0.822288 kJ

Step 2:
q2 = m * ΔH
q2 = 8.88 g * 0.335 kJ/g
q2 = 2.9708 kJ

Step 3:
Total heat = q1 + q2
Total heat = 0.822288 kJ + 2.9708 kJ
Total heat = 3.793088 kJ

Therefore, the minimum amount of heat needed to convert the ice sample to liquid water is approximately 3.793 kJ.