how much ice at -10degree celsius is needed to cool .1 kg of ethyl alcohol and .1kg of water from 20 degree celcius to 5 degree celsius?
The amount of ice needed to cool .1 kg of ethyl alcohol and .1 kg of water from 20 degree Celsius to 5 degree Celsius is approximately 0.2 kg. This is because the specific heat capacity of ethyl alcohol is 2.44 kJ/kgK and the specific heat capacity of water is 4.18 kJ/kgK. Therefore, the total amount of energy needed to cool the two substances from 20 degree Celsius to 5 degree Celsius is (2.44 x 0.1 x 15) + (4.18 x 0.1 x 15) = 4.62 kJ. Since the latent heat of fusion of ice is 334 kJ/kg, the amount of ice needed to cool the two substances is 4.62/334 = 0.0138 kg, which is approximately 0.2 kg.
To calculate the amount of ice needed to cool the ethyl alcohol and water, we need to consider the heat transfer between the substances.
Step 1: Calculate the heat gained or lost by each substance using the formula:
Q = m * c * ΔT
where:
Q is the heat gained or lost by the substance,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
For ethyl alcohol:
Q_ethanol = m_ethanol * c_ethanol * ΔT_ethanol
Where:
m_ethanol = 0.1 kg (mass of ethyl alcohol)
c_ethanol = 2.44 J/g°C (specific heat capacity of ethyl alcohol)
ΔT_ethanol = (5°C - 20°C) = -15°C (change in temperature)
Q_ethanol = 0.1 kg * 2.44 J/g°C * -15°C
For water:
Q_water = m_water * c_water * ΔT_water
Where:
m_water = 0.1 kg (mass of water)
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = (5°C - 20°C) = -15°C (change in temperature)
Q_water = 0.1 kg * 4.18 J/g°C * -15°C
Step 2: Calculate the heat released by melting the ice using the formula:
Q_ice = m_ice * ΔH_fusion
Where:
Q_ice is the heat released by the ice,
m_ice is the mass of the ice, and
ΔH_fusion is the heat of fusion of ice (334 J/g).
Since the ice is at -10°C and we need to melt it to 0°C:
m_ice = Q_ice / ΔH_fusion
Step 3: Calculate the total heat needed to cool both substances and melt the ice:
Q_total = Q_ethanol + Q_water + Q_ice
Now, let's calculate the values:
Q_ethanol = 0.1 kg * 2.44 J/g°C * -15°C = -36.6 J
Q_water = 0.1 kg * 4.18 J/g°C * -15°C = -62.7 J
Q_ice = m_ice * ΔH_fusion
m_ice = Q_ice / ΔH_fusion = (Q_ethanol + Q_water) / ΔH_fusion
Q_total = Q_ethanol + Q_water + Q_ice
Now, let's substitute the values and calculate:
Q_ice = (Q_ethanol + Q_water) / ΔH_fusion
Q_ice = (-36.6 J + (-62.7 J)) / 334 J/g = -99.3 J / 334 J/g ≈ -0.297 g
The negative sign indicates that heat is released while melting the ice.
Therefore, approximately 0.297 g of ice is needed to cool 0.1 kg of ethyl alcohol and 0.1 kg of water from 20°C to 5°C.
To find out how much ice is needed to cool the ethyl alcohol and water, you need to calculate the heat transferred during the process. The amount of heat transferred can be determined using the specific heat capacities and temperature changes of the substances involved.
Let's break down the problem into a few steps:
1. Calculate the heat transfer for ethyl alcohol:
The specific heat capacity of ethyl alcohol is 2.44 J/g°C. The mass of ethyl alcohol is 0.1 kg, and the temperature change is from 20°C to 5°C.
Q_ethyl = mass_ethyl * specific_heat_ethyl * change_in_temperature_ethyl
Q_ethyl = 0.1 kg * 2.44 J/g°C * (5°C - 20°C)
2. Calculate the heat transfer for water:
The specific heat capacity of water is 4.18 J/g°C. The mass of water is also 0.1 kg, and the temperature change is from 20°C to 5°C.
Q_water = mass_water * specific_heat_water * change_in_temperature_water
Q_water = 0.1 kg * 4.18 J/g°C * (5°C - 20°C)
3. Sum up the heat transfers:
Since the heat transfer during cooling comes from the ice, the total heat transferred should be equal to the heat absorbed by ice, which can then be calculated as follows:
Q_total = Q_ethyl + Q_water
4. Calculate the heat absorbed by ice:
To melt ice into water, it requires 333.55 J/g of heat.
Q_ice = Q_total / 333.55 J/g
Based on these calculations, you can determine the amount of ice required to cool both the ethyl alcohol and water from 20°C to 5°C.