If 3.20 kcal of heat is added to 1.00 kg of ice at 0 degrees C, how much water at 0 degrees C is produced, and how much ice remains?
Could someone post the answer n the equation?
Sure, I'd be happy to help! But before we get to the answer, let me just say, if you add heat to ice, things can get pretty steamy! Now, back to business.
Let's break this down step by step. To turn ice into water, we need to provide enough energy to melt it. This energy is known as the heat of fusion. For water, this value is approximately 334 J/g.
First, let's convert the given heat value into joules. Since 1 kcal = 4184 J, we have 3.20 kcal = 3.20 x 4184 J = 13,388.8 J.
To find out how much ice melts, we can divide the energy added by the heat of fusion:
Mass of ice melted = Energy added / Heat of fusion
Mass of ice melted = 13,388.8 J / 334 J/g
Now, let's convert the mass from grams to kilograms for convenience:
Mass of ice melted = (13,388.8 J / 334 J/g) / 1000 g/kg
Simplifying this equation would give you the mass of water produced.
Now, to find out how much ice remains, you can subtract the mass of water produced from the initial mass of ice (1.00 kg) to get the final mass of ice.
I won't bore you with all the calculations, but if you plug in the numbers, you should be able to get the final answer! Just be patient, because solving equations can sometimes be a bit of a melting journey.
To determine the amount of water at 0 degrees Celsius produced and the amount of ice remaining after adding 3.20 kcal of heat to 1.00 kg of ice at 0 degrees Celsius, we need to consider the heat required for each phase change.
The equation for calculating the heat involved in a phase change is:
Q = m × ΔH
Where:
Q is the heat transferred
m is the mass of the substance
ΔH is the heat of fusion (for melting) or heat of solidification (for freezing)
For ice to water, we have:
Q1 = m × ΔH_fusion
And for water to ice, we have:
Q2 = m × ΔH_solidification
First, let's calculate the heat required to melt the ice into water. The heat of fusion of ice is 334 kJ/kg.
Q1 = (1.00 kg) × (334 kJ/kg)
Q1 = 334 kJ
Since 1 kcal is equal to 4.18 kJ, we can convert Q1 into kcal:
Q1 = 334 kJ × (1 kcal/4.18 kJ)
Q1 = 80.1 kcal
By subtracting this amount of heat from the initial 3.20 kcal, we can determine the amount of heat remaining to freeze the water into ice.
Remaining heat = 3.20 kcal - 80.1 kcal
Remaining heat = -76.9 kcal (negative sign indicates heat will be released during freezing)
Now, let's calculate the mass of water at 0 degrees Celsius produced:
Q2 = (m_water) × (ΔH_solidification)
Since the heat of solidification (freezing) is the negative of the heat of fusion (melting), we can use the same value of 334 kJ/kg but with a negative sign to indicate heat release.
Q2 = (m_water) × (-334 kJ/kg)
Q2 = (m_water) × (-334 kJ) (as we're looking for mass)
Converting the remaining heat into kJ:
Remaining heat = -76.9 kcal × 4.18 kJ/kcal
Remaining heat = -321 kJ
Now, we can equate Q2 with the remaining heat:
Remaining heat = (m_water) × (-334 kJ)
-321 kJ = (m_water) × (-334 kJ)
Solving for m_water:
m_water = -321 kJ / -334 kJ
m_water ≈ 0.96 kg
Finally, we can determine the mass of ice remaining:
m_ice = m_initial - m_water
m_ice = 1.00 kg - 0.96 kg
m_ice = 0.04 kg
Therefore, approximately 0.96 kg of water at 0 degrees Celsius is produced, and approximately 0.04 kg of ice remains.
To solve this problem, we can use the concept of heat transfer and the specific heat capacity of water and ice.
The equation we can use is:
Q = mcΔT
Where:
- Q is the heat transferred (in calories or joules)
- m is the mass of the substance (in grams or kilograms)
- c is the specific heat capacity of the substance (in calories/gram°C or joules/gram°C)
- ΔT is the change in temperature (in °C)
Let's break down the problem step by step:
1. Determine the heat required to raise the temperature of ice at 0°C to water at 0°C.
To do this, we need to calculate the heat absorbed by the ice:
Q1 = m1 * c1 * ΔT1
The mass of the ice is given as 1.00 kg, which is equal to 1000 grams (m1 = 1000 g).
The specific heat capacity for ice is approximately 2.09 J/g°C (c1 = 2.09 J/g°C).
The change in temperature (ΔT1) is from 0°C to 0°C, so there is no change (ΔT1 = 0°C).
Plugging these values into the equation, we get:
Q1 = 1000 g * 2.09 J/g°C * 0°C = 0 J
This means no heat is absorbed by the ice because it does not change its temperature.
2. Determine the heat required to melt the ice into water.
To do this, we need to calculate the heat required for the phase change from solid to liquid:
Q2 = m2 * ΔHf
The mass of the ice that needs to be melted is still 1.00 kg, or 1000 grams (m2 = 1000 g).
The heat of fusion for water is 333.55 J/g.
Plugging these values into the equation, we get:
Q2 = 1000 g * 333.55 J/g = 333550 J
3. Determine the heat required to raise the temperature of water at 0°C to the final temperature.
To do this, we need to calculate the heat absorbed by the water:
Q3 = m3 * c3 * ΔT3
The mass of the water we are trying to find is denoted as m3.
The specific heat capacity of water is approximately 4.18 J/g°C (c3 = 4.18 J/g°C).
The change in temperature (ΔT3) is also from 0°C to 0°C, so there is no change (ΔT3 = 0°C).
Plugging these values into the equation, we get:
Q3 = m3 * 4.18 J/g°C * 0°C = 0 J
This means no additional heat is required to raise the temperature of water at 0°C.
4. Calculate the total heat required to convert ice to water at 0°C:
Q_total = Q1 + Q2 + Q3
= 0 J + 333550 J + 0 J
= 333550 J
5. Convert the total heat from joules to kilocalories:
1 kilocalorie (kcal) = 4184 joules (J)
Q_total_kcal = Q_total / 4184 J/kcal
= 333550 J / 4184 J/kcal
≈ 79.76 kcal
Therefore, if 3.20 kcal of heat is added, and the total heat required to convert ice to water at 0°C is 79.76 kcal, we can subtract the two values to find the remaining ice:
Remaining ice = Total heat required - Heat added
= 79.76 kcal - 3.20 kcal
≈ 76.56 kcal
Thus, approximately 76.56 kcal of ice remains.