How much heat is necessary to change 345 g of ice at -11°C to water at 20°C?
Well, that's an icy situation you've got there! To calculate the heat needed to change the ice to water, we need to consider two processes: first we need to warm up the ice from -11°C to 0°C, and then we need to melt it into water at 0°C. Shall we do some clown calculations?
To warm up the ice from -11°C to 0°C, we need to find the heat using the specific heat capacity of ice. But hey, don't worry, I'm not going to give you a cold response - it's approximately 2.09 J/g°C. So, we multiply the mass of the ice (345 g) by the temperature change (0°C - (-11°C) = 11°C) and the specific heat capacity (2.09 J/g°C):
Heat = 345 g * 11°C * 2.09 J/g°C
Got an ice calculator? Great! Let's move on.
Now let's melt the ice into water at 0°C. The heat needed to melt ice is called the heat of fusion. And what do you know, the heat of fusion for ice is approximately 334 J/g. Simply multiply the mass of the ice (345 g) by the heat of fusion (334 J/g):
Heat = 345 g * 334 J/g
Now, to find the total heat needed, we add the heat required to warm up the ice to the heat needed to melt it:
Total Heat = Heat to warm up ice + Heat to melt ice
Feel free to do the math and find out the grand total. Just be careful not to get cold feet during the calculations!
To calculate the amount of heat required to change the ice at -11°C to water at 20°C, we need to consider two separate processes:
1. Heating the ice to its melting point.
2. Melting the ice and further heating the resulting water to the final temperature.
Let's break down these steps:
Step 1: Heating the ice to 0°C
To heat the ice from -11°C to 0°C, we need to calculate the amount of heat required using the formula:
Q1 = m * Cp * ΔT
Where:
Q1 is the heat required
m is the mass of the ice (345 g)
Cp is the specific heat capacity of ice (2.09 J/g°C)
ΔT is the change in temperature (0°C - (-11°C) = 11°C)
Plugging in the values:
Q1 = 345 g * 2.09 J/g°C * 11°C
Q1 ≈ 8,107.95 J (rounded to four decimal places)
Step 2: Melting the ice and heating the water
To calculate the heat required to melt the ice and heat the resulting water to 20°C, we need to consider two separate processes:
a. Melting the ice
To melt the ice at 0°C, the heat required can be calculated using the formula:
Q2a = m * ΔHf
Where:
Q2a is the heat required for melting
m is the mass of the ice (345 g)
ΔHf is the enthalpy of fusion for water (334 J/g)
Plugging in the values:
Q2a = 345 g * 334 J/g
Q2a ≈ 115,230 J (rounded to four decimal places)
b. Heating the water
To heat the resulting water from 0°C to 20°C, we use the formula:
Q2b = m * Cp * ΔT
Where:
Q2b is the heat required for heating the water
m is the mass of the water (345 g)
Cp is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (20°C - 0°C = 20°C)
Plugging in the values:
Q2b = 345 g * 4.18 J/g°C * 20°C
Q2b ≈ 28,689 J (rounded to four decimal places)
Adding up the heat required in both steps:
Q2 = Q2a + Q2b
Q2 ≈ 115,230 J + 28,689 J
Q2 ≈ 143,919 J (rounded to four decimal places)
Therefore, the total amount of heat necessary to change 345 g of ice at -11°C to water at 20°C is approximately 152,027.95 J (Q1 + Q2).
To find out how much heat is necessary to change a substance from one state to another, you need to calculate the amount of heat required for each phase transition (melting/freezing) and for raising or lowering the temperature.
To solve this problem, we need to consider the following steps:
1. Calculate the heat required to bring the ice to its melting point (0°C).
The heat required for this step can be found using the formula: Q = m * C * ΔT,
where Q is the heat (in Joules), m is the mass of the substance (in grams), C is the specific heat capacity of the substance (in J/g·°C), and ΔT is the change in temperature.
Given: m = 345 g, C = 2.09 J/g·°C, and ΔT = (0°C - (-11°C)) = 11°C
Q1 = 345 g * 2.09 J/g·°C * 11°C = 8051.95 J
2. Calculate the heat required to melt the ice.
The heat required for this step can be given by the formula: Q = m * ΔHf,
where Q is the heat (in Joules), m is the mass of the substance (in grams), and ΔHf is the heat of fusion (molar enthalpy change) for the substance (in J/g).
Given: m = 345 g and ΔHf = 333.55 J/g
Q2 = 345 g * 333.55 J/g = 115030.75 J
3. Calculate the heat required to raise the water temperature from 0°C to 20°C.
Using the same formula as in step 1, we can find: Q = m * C * ΔT.
Given: m = 345 g, C = 4.18 J/g·°C, and ΔT = (20°C - 0°C) = 20°C
Q3 = 345 g * 4.18 J/g·°C * 20°C = 28869 J
4. Add up the heat required for each step to get the total heat required.
Total heat required = Q1 + Q2 + Q3 = 8051.95 J + 115030.75 J + 28869 J = 152951.70 J
Therefore, the total heat necessary to change 345 g of ice at -11°C to water at 20°C is approximately 152,951.70 Joules.