An insulated Thermos contains 140 cm3 of hot coffee at 89.0°C. You put in a 11.0 g ice cube at its melting point to cool the coffee. By how many degrees (in Celsius) has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment. The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm3.
http://www.jiskha.com/display.cgi?id=1323003502
To find the change in temperature of the coffee, we need to calculate the amount of heat transferred from the coffee to melt the ice cube.
First, let's calculate the amount of heat required to melt the ice cube:
1. Calculate the mass of the ice cube:
Mass = 11.0 g
2. Convert mass to kilograms:
Mass = 11.0 g ÷ 1000 = 0.011 kg
3. Calculate the heat required to melt the ice:
Heat = Mass × Latent heat of fusion
Heat = 0.011 kg × 333,000 J/kg
Heat = 3663 J
Now, let's calculate the change in temperature of the coffee:
1. Calculate the initial heat of the coffee:
Initial heat = Initial mass × Specific heat capacity × Initial temperature
Note: We need to convert the volume of the coffee to mass using its density.
Density of water = 1.00 g/cm³
Volume of coffee = 140 cm³
Mass of coffee = Volume × Density
Mass of coffee = 140 cm³ × 1.00 g/cm³
Mass of coffee = 140 g
Converting grams to kilograms:
Mass of coffee = 140 g ÷ 1000 = 0.14 kg
Initial heat = 0.14 kg × 4186 J/kg·K × 89.0°C
Initial heat = 52985.64 J
2. Calculate the final heat of the coffee:
Final heat = Initial heat - Heat
Final heat = 52985.64 J - 3663 J
Final heat = 49322.64 J
3. Calculate the final temperature of the coffee:
Final temperature = Final heat ÷ (Mass × Specific heat capacity)
Final temperature = 49322.64 J ÷ (0.14 kg × 4186 J/kg·K)
Final temperature ≈ 24.75°C
Therefore, the coffee cools by approximately 89.0°C - 24.75°C = 64.25°C once the ice cube has melted and equilibrium is reached.
To determine the change in temperature of the coffee once the ice has melted and equilibrium is reached, we must consider both the cooling of the coffee and the melting of the ice.
First, let's calculate the heat lost by the coffee as it cools down. The amount of heat lost can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat lost (in joules)
m is the mass of the coffee (in kilograms)
c is the specific heat capacity of water (4186 J/kg·K)
ΔT is the change in temperature of the coffee (in Kelvin)
We know that the final temperature of the coffee is 0°C, since it will reach equilibrium with the melted ice. Therefore, the change in temperature of the coffee is:
ΔT = 0°C - 89.0°C = -89.0 K
Next, we need to calculate the mass of the coffee. We are given the volume of the coffee and the density of water. Since the density of water is 1.00 g/cm3, and the volume of the coffee is 140 cm3, the mass can be calculated as:
m = density * volume
= 1.00 g/cm3 * 140 cm3
= 140 g
= 0.140 kg
Now we can calculate the heat lost by the coffee:
Q = mcΔT
= 0.140 kg * 4186 J/kg·K * -89.0 K
= -5232.32 J
Now, let's calculate the heat gained by the ice as it melts. The heat gained can be calculated using the formula:
Q = mL
Where:
Q is the heat gained (in joules)
m is the mass of the ice (in kilograms)
L is the latent heat of fusion (333 kJ/kg or 333000 J/kg)
The heat gained by the ice is equal to the heat lost by the coffee. Therefore:
mL = -5232.32 J
To find the mass of the ice, we divide both sides of the equation by L:
m = -5232.32 J / 333000 J/kg
= -0.0157 kg (since mass cannot be negative, ignore the negative sign)
Therefore, the mass of the ice is approximately 0.0157 kg.
Next, let's find the change in temperature of the ice as it melts. The change in temperature can be calculated using the formula:
ΔT = Q / (m * c)
Where:
Q is the heat gained by the ice (in joules)
m is the mass of the ice (in kilograms)
c is the specific heat capacity of water (4186 J/kg·K)
ΔT = -5232.32 J / (0.0157 kg * 4186 J/kg·K)
= -78.5 K
Since the ice is at its melting point, the change in temperature is equal to the decrease in temperature of the coffee:
ΔT = -78.5 K
Therefore, the coffee has cooled by 78.5°C once the ice has melted and equilibrium is reached.