a 9.0 g ice cube at -10 C is put into a thermos flask containing 100 cm^3 of water at 20 C. by how much has the entropy of the cube-water system changed when the equilibrium is reached? the specific heat of ice is 2220 J/kg K.
Add the entropy gains of the ice cube (as it heats up to the equilibrium T) and the cooling water (as is cools down), The original liquid water actually loses entropy, but the ice gains more.
You will need to calculate the equilibrium temperature first. All of the ice will melt in this case.
The water-cooling and ice-heating steps will require an integration of dT/T. The melting step will be isentropic.
The heating of the melted ice to the final equilibrium temperature also has an associated entropy gain that must be included.
To calculate the change in entropy of the cube-water system when equilibrium is reached, we need to consider the change in entropy of the ice cube and the change in entropy of the water separately.
1. Change in entropy of the ice cube:
The change in entropy of the ice cube can be calculated using the formula:
ΔS_ice = q_ice / T_ice
where:
ΔS_ice is the change in entropy of the ice cube (in J/K)
q_ice is the heat transfer of the ice cube (in J)
T_ice is the temperature of the ice cube (in K)
First, let's calculate the heat transfer of the ice cube using the formula:
q_ice = m_ice * c_ice * ΔT_ice
where:
m_ice is the mass of the ice cube (in kg)
c_ice is the specific heat of ice (in J/kg K)
ΔT_ice is the change in temperature of the ice cube (final temperature - initial temperature) (in K)
Given:
m_ice = 9.0 g = 0.009 kg
c_ice = 2220 J/kg K
ΔT_ice = 0 K - (-10 K) = 10 K
q_ice = 0.009 kg * 2220 J/kg K * 10 K
q_ice = 199.8 J
Now, we can calculate the change in entropy of the ice cube:
ΔS_ice = 199.8 J / 263 K
ΔS_ice ≈ 0.76 J/K
2. Change in entropy of the water:
The change in entropy of the water can be calculated using the formula:
ΔS_water = q_water / T_water
where:
ΔS_water is the change in entropy of the water (in J/K)
q_water is the heat transfer of the water (in J)
T_water is the temperature of the water (in K)
Since the water is in thermal equilibrium, no heat transfer occurs. Therefore, q_water = 0 J.
Thus, the change in entropy of the water is zero:
ΔS_water = 0 J/K
Finally, to find the change in entropy of the cube-water system, we sum up the changes in entropy of the ice cube and the water:
ΔS_system = ΔS_ice + ΔS_water
ΔS_system ≈ 0.76 J/K + 0 J/K
ΔS_system ≈ 0.76 J/K
Therefore, the entropy of the cube-water system has changed by approximately 0.76 J/K when equilibrium is reached.
To find the change in entropy of the cube-water system, we need to consider the change in entropy of both the ice cube and the water separately. The change in entropy for each component can be calculated using the formula:
ΔS = mcΔT
Where:
ΔS = Change in entropy
m = Mass of the component (in kg)
c = Specific heat capacity of the component (in J/kg K)
ΔT = Change in temperature (in K)
Let's calculate the change in entropy for the ice cube first:
1. Convert the mass of the ice cube from grams to kilograms:
Mass of the ice cube = 9.0 g = 0.009 kg
2. Calculate the change in temperature for the ice cube:
ΔT = final temperature - initial temperature
Since the ice cube is initially at -10°C and the equilibrium temperature is reached at 0°C, the change in temperature is:
ΔT = (0°C - (-10°C)) = 10 K
3. Calculate the change in entropy for the ice cube:
ΔS_ice = m_ice * c_ice * ΔT
ΔS_ice = 0.009 kg * 2220 J/kg K * 10 K
Now, let's calculate the change in entropy for the water:
1. Calculate the change in temperature for the water:
ΔT = final temperature - initial temperature
Since the water is initially at 20°C and the equilibrium temperature is reached at 0°C, the change in temperature is:
ΔT = (0°C - 20°C) = -20 K
2. Calculate the change in entropy for the water:
ΔS_water = m_water * c_water * ΔT
The mass of the water can be calculated from its density, which is approximately 1 g/cm^3 (or 1000 kg/m^3):
Volume of the water = 100 cm^3 = 0.1 L = 0.1 kg
ΔS_water = 0.1 kg * c_water * (-20 K)
Finally, to find the total change in entropy for the cube-water system, we add the values obtained for the ice cube and the water:
ΔS_total = ΔS_ice + ΔS_water
Simplifying the equation using the calculated values will give you the final answer for the change in entropy of the cube-water system.