(a) In a hot summer afternoon, you just finished a football game. You rushed back to your home to grab a coke to cool yourself down. Unfortunately, you found out that you forgot to store your coke in the
refrigerator but placed it on the dining table. Assuming that the coke weighs 300 g and is initially at 25 degree C, how much 0 degree C ice do you need to add to the coke such that the coke will drop to 10 degree C? (Specific
heat capacity of water = 4200 Jkg^(-1) degree C^(-1), specific latent heat of fusion of ice = 3.34×10^(5) Jkg^(-1), specific heat capacity of coke = 4000 Jkg^(-1) degree C^(-1)
(b) A beaker containing 100 g of mysterious substance X initially at temperature 29 degree C is placed on a heater. The rate of energy transfer to X is 10 W. Sketch a graph to show how does the temperature of X change with time for 30 minutes. (Specific heat capacity of X in solid form = 1295 Jkg^(-1) degree C^(-1), specific latent heat of fusion of X = 4.34×10^(4) Jkg^(-1), specific heat capacity of X in liquid form = 2362 Jkg^(-1 )degree C^(-1), melting point of X is 45 degree C and boiling point of X is 235 degree C)
(c) In a sealed container of size 500 cm^3, the number of air molecules inside is 7×10^(21) and the temperature
is 32 degree C? What is the pressure exerted by the gas on the container?
(a) To solve this problem, we need to consider the energy transfer that occurs when the ice melts and then raises the temperature of the coke.
First, we need to find the amount of heat energy required to decrease the temperature of the coke from 25°C to 10°C. We can use the formula:
Q = mcΔT
Where:
Q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
For the coke, the mass is given as 300g and the specific heat capacity as 4000 J/kg°C. Thus:
Q(coke) = (300g)(4000 J/kg°C)(10°C-25°C)
Next, we need to find the amount of heat energy released when the ice melts. This can be calculated using the formula:
Q = mLf
Where:
Q = heat energy
m = mass
Lf = specific latent heat of fusion
For the ice, the specific latent heat of fusion is given as 3.34×10^5 J/kg. We need to find the mass of the ice required to release the same amount of heat energy as absorbed by the coke. Let's assume the mass of ice needed is mi:
Q(coke) = Q(ice)
(300g)(4000 J/kg°C)(10°C-25°C) = mi(3.34×10^5 J/kg)
Now we can solve for mi:
mi = [(300g)(4000 J/kg°C)(10°C-25°C)] / (3.34×10^5 J/kg)
Once we find the mass of ice needed, we can convert it to grams (g) if required.
(b) To sketch a graph showing how the temperature of substance X changes with time for 30 minutes, we need to consider the three stages of heating: solid heating, fusion, and liquid heating.
Since the power input is constant, the rate of energy transfer (P) is given as 10W. We can calculate the total heat energy transferred during each stage to determine the temperature changes.
In the solid heating stage, we can use the formula:
Q = mcΔT
Where:
Q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
For the solid heating stage, the mass is given as 100g, and the specific heat capacity is given as 1295 J/kg°C. We need to find the change in temperature (ΔT) by rearranging the formula:
ΔT = Q / (mc)
For the fusion stage, the heat energy required to melt the substance is given by:
Q(fusion) = mLf
Where:
Q(fusion) = heat energy
m = mass
Lf = specific latent heat of fusion
For the fusion stage, the mass is still 100g, and the specific latent heat of fusion is given as 4.34×10^4 J/kg.
In the liquid heating stage, we again use the formula:
Q = mcΔT
Where:
Q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
For the liquid heating stage, the mass is still 100g, but the specific heat capacity changes to 2362 J/kg°C. We need to find the change in temperature (ΔT) using the same rearranged formula as before:
ΔT = Q / (mc)
Once we have calculated the changes in temperature for each stage, we can plot them on a graph with time on the x-axis and temperature on the y-axis. We can show the three stages as distinct segments on the graph.
(c) To calculate the pressure exerted by the gas on the container, we can use the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, we need to convert the volume from cm^3 to m^3. Since 1 m^3 = 10^6 cm^3, we have:
V = 500 cm^3 = 500 / (10^6) m^3
Next, we need to calculate the number of moles using the ideal gas law, rearranged as:
n = PV / (RT)
Given that the number of air molecules is 7×10^21, we can convert this to moles using Avogadro's number (6.022×10^23 molecules/mol):
n = (7×10^21 molecules) / (6.022×10^23 molecules/mol)
Finally, we can substitute these values into the ideal gas law to find the pressure:
P = (nRT) / V
With the given temperature of 32°C, we need to convert it to Kelvin by adding 273.15:
T = 32°C + 273.15
Substituting all the values into the equation will give us the pressure exerted by the gas on the container.