A 50g ice cube is added to 110g of water in a 62g aluminum cup. The water and cup initially have a temperature of 23°C. Find the initial temperature of the ice cube such that all of the water is just turned to ice. Use the following specific heats: water = 4186 J/(kg•K), ice = 2090 J/(kg•K), and aluminum = 900 J/(kg•K). The latent heat of fusion for water is 33.5e4 J/kg
find the heat needed...
Heatneeded=masswater*c(0-23)-masswater*Lf+massAl*Cal*(0-23)
Notice this heat is "negative), meaning it was lost. (it is negative heat gained).
now, the heat the ice gained.
massice*Cice*(0-Ti)
Now add the heats gained...set to zero.
heat ice gained+heat water(and aluminum) gaied [remember this is negative] =0
solve for Ti
97.93 celcius
To solve this problem, we need to consider the energy transferred between the ice cube, water, and aluminum cup. The amount of energy transferred is equal to the change in thermal energy of each substance, which can be calculated using the specific heat capacity equation:
Q = mcΔT
where Q is the energy transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the energy transferred to the water:
Q_water = m_water * c_water * ΔT_water
where m_water = 110g (converted to kg),
c_water = 4186 J/(kg•K),
and ΔT_water = 0°C - 23°C = -23K.
Q_water = 0.110kg * 4186 J/(kg•K) * -23K
Q_water = -10799.32 J
The negative sign indicates that the water is losing energy.
Next, let's calculate the energy transferred to the aluminum cup:
Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
where m_aluminum = 62g (converted to kg),
c_aluminum = 900 J/(kg•K),
and ΔT_aluminum = 0°C - 23°C = -23K.
Q_aluminum = 0.062kg * 900 J/(kg•K) * -23K
Q_aluminum = -1281.6 J
Again, the negative sign indicates that the aluminum cup is losing energy.
Now, let's calculate the energy required to convert the water to ice:
Q_ice = m_ice * L_fusion
where m_ice = 50g (converted to kg),
L_fusion = 33.5e4 J/kg.
Q_ice = 0.050kg * 33.5e4 J/kg
Q_ice = 167500 J
The positive sign indicates that energy is added to the system to convert the water to ice.
The total energy transferred is the sum of the energy transferred to the water, aluminum, and the energy required to freeze the water:
Q_total = Q_water + Q_aluminum + Q_ice
Q_total = -10799.32 J + (-1281.6 J) + 167500 J
Q_total = 155419.08 J
To find the initial temperature of the ice cube such that all of the water is just turned to ice, we need to consider the change in thermal energy of the ice cube. Since no heat is transferred to or from the ice cube initially, we can use the equation:
Q_icecube = m_icecube * c_ice * ΔT_icecube
where Q_icecube = 0 J, as no energy is transferred,
m_icecube = 50g (converted to kg),
c_ice = 2090 J/(kg•K),
and ΔT_icecube is the change in temperature of the ice cube.
So, we can solve for ΔT_icecube:
0 = 0.050kg * 2090 J/(kg•K) * ΔT_icecube
ΔT_icecube = 0
Therefore, the initial temperature of the ice cube is 0°C or 273.15K.
Note: The initial temperature of the ice cube is typically taken as its melting point, which is 0°C. This occurs because the ice cube is already at its freezing point, and no further energy transfer is needed to lower its temperature.