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.
At equilbrium, the melted ice (mass m) and the water it changes into will gain as much heat as the original coffee (mass M) loses.
0.140g*(89 - T)*4186
= 0.011*[333,000 + T*4186]
52,160 -586 T = 3663 + 46 T
632 T = 48,497
T = 77 C
Solve for final temperature T
Heat releasing from coffee (water in this case) is equal to the heat gain for ice melting to water at 0 deg plus water to raise for the final temp.
To solve this problem, we need to calculate the amount of heat transferred from the hot coffee to melt the ice cube and then the resulting change in temperature.
Step 1: Calculate the heat transferred to melt the ice cube.
The latent heat of fusion is the amount of heat required to convert a substance from solid to liquid phase at its melting point. In this case, it is 333 kJ/kg.
The mass of the ice cube is 11.0 g.
Therefore, the total heat transferred to melt the ice cube can be calculated by multiplying the mass of the ice cube by the latent heat of fusion:
Heat transferred = Mass of ice cube × Latent heat of fusion
Heat transferred = 11.0 g × (333 kJ/kg × 1000 g/kg)
Heat transferred = 11.0 g × 333000 J/kg
Heat transferred = 3.663 x 10^6 J
Step 2: Calculate the change in temperature of the coffee.
The specific heat capacity of water is 4186 J/kg·K.
To determine the change in temperature, we can use the formula:
Heat transferred = Mass of coffee × Specific heat capacity of water × Change in temperature
Since we know the heat transferred (3.663 x 10^6 J) and the mass of the coffee is unknown, we need to find the mass of the coffee to determine the change in temperature.
The density of water is 1.00 g/cm³, which is equivalent to 1.00 kg/L or 1000 kg/m³.
To find the mass of the coffee, we use the formula:
Mass of coffee = Volume of coffee × Density of water
Mass of coffee = 140 cm³ × (1.00 g/cm³ × 1 kg/1000 g)
Mass of coffee = 140 cm³ × 0.001 kg/cm³
Mass of coffee = 0.14 kg
Now we can calculate the change in temperature:
3.663 x 10^6 J = 0.14 kg × 4186 J/kg·K × Change in temperature
Change in temperature = (3.663 x 10^6 J) / (0.14 kg × 4186 J/kg·K)
Change in temperature = 61.69 K
Step 3: Convert the change in temperature from Kelvin to Celsius.
To convert the change in temperature from Kelvin to Celsius, we subtract 273.15 from the value obtained:
Change in temperature in Celsius = 61.69 K - 273.15
Change in temperature in Celsius = -211.46°C (approximately)
Therefore, the coffee has cooled by approximately 211.46°C once the ice has melted and equilibrium is reached.
To determine the number of degrees by which the coffee has cooled after the ice melts and equilibrium is reached, we need to consider the energy gained by the ice as it melts and the energy lost by the coffee.
First, let's calculate the energy gained by the ice as it melts.
The specific heat formula is given by:
Q = m * c * ΔT
Where:
Q - Energy gained (or lost)
m - Mass of the substance
c - Specific heat
ΔT - Change in temperature
Since the ice is at its melting point, there is no change in temperature, and ΔT = 0 °C.
The mass of the ice is equal to its weight divided by the density:
m_ice = 11.0 g / (1.00 g/cm³) = 11.0 cm³
Now, we can calculate the energy gained by the ice when it melts:
Q_ice = m_ice * c_ice * ΔT_ice
Since ΔT_ice = 0 °C, Q_ice = 0 J
Next, we need to calculate the energy lost by the coffee.
For the coffee, the energy lost can be determined using:
Q_coffee = m_coffee * c_water * ΔT_coffee
Since there is no energy exchange with the environment, the energy gained by the ice must be equal to the energy lost by the coffee:
Q_coffee = -Q_ice
m_coffee * c_water * ΔT_coffee = -m_ice * c_ice * ΔT_ice
Rearranging the equation, we find:
ΔT_coffee = (-m_ice * c_ice * ΔT_ice) / (m_coffee * c_water)
Substituting the given values:
ΔT_coffee = (-11.0 g * 333 kJ/kg) / (140 cm³ * 1.00 g/cm³ * 4186 J/kg·K)
Converting the units:
ΔT_coffee = (-11.0 g * 333000 J/kg) / (140 cm³ * 1.00 g/cm³ * 4186 J/kg·K)
Simplifying the expression:
ΔT_coffee ≈ -7.1 °C
Therefore, the coffee has cooled by approximately 7.1 °C once the ice has melted and equilibrium is reached.