A thermos contains 152 cm3 of coffee at 93.8 °C. To cool the coffee, you drop two 11.6-g ice cubes into the thermos. The ice cubes are initially at 0 °C and melt completely. What is the final temperature of the coffee in degrees Celsius? Treat the coffee as if it were water.
Calculate the energy required to warm up 25kg of bath water at 10 degrees celsius to a temperature of 20 degrees celsius?
To find the final temperature of the coffee, we can use the principle of conservation of energy. The amount of heat lost by the coffee will be equal to the amount of heat gained by the ice cubes.
First, let's determine the heat lost by the coffee. We can use the specific heat capacity formula:
Q = mcΔT
Where:
- Q represents the heat lost by the coffee (in joules)
- m is the mass of the coffee (in grams)
- c is the specific heat capacity of water (4.18 J/g°C)
- ΔT is the change in temperature (final temperature - initial temperature)
The mass of the coffee can be found using the density formula:
density = mass/volume
The density of coffee is approximately equal to the density of water, which is 1 g/cm³. We have the volume of the coffee, which is 152 cm³, so we can find the mass of the coffee:
mass = density * volume
The mass of the coffee is 1 g/cm³ * 152 cm³ = 152 g.
Next, we calculate the amount of heat lost by the coffee. The initial temperature of the coffee is 93.8 °C, and we need to find the final temperature.
Q = mcΔT
Q = 152 g * 4.18 J/g°C * (final temperature - 93.8 °C)
Now, let's determine the heat gained by the ice cubes. Since the ice cubes are initially at 0 °C and they melt completely, we need to consider the heat required to raise their temperature from 0 °C to the melting point of ice and then the heat necessary to melt the ice completely.
The specific heat capacity of ice is also 4.18 J/g°C. We have two ice cubes, each weighing 11.6 g, so the total mass of the ice is 2 * 11.6 g = 23.2 g.
Heat required to raise the temperature of ice from 0 °C to the melting point of ice:
q1 = mcΔT
q1 = 23.2 g * 4.18 J/g°C * (0 °C - 0 °C) = 0 J
Heat required to melt the ice completely:
q2 = mL
q2 = 23.2 g * 334 J/g = 7732.8 J
Since the heat gained by the ice cubes is equal to the heat lost by the coffee, we can set up an equation and solve for the final temperature:
Q = Q + Q
152 g * 4.18 J/g°C * (final temperature - 93.8 °C) = 0 J + 7732.8 J
152 g * 4.18 J/g°C * (final temperature - 93.8 °C) = 7732.8 J
Now we solve for the final temperature:
(152 g * 4.18 J/g°C * final temperature) - (152 g * 4.18 J/g°C * 93.8 °C) = 7732.8 J
627.36 g°C * final temperature - 60250.56 g°C = 7732.8 J
627.36 g°C * final temperature = 60250.56 g°C + 7732.8 J
final temperature = (60250.56 g°C + 7732.8 J) / 627.36 g°C
final temperature ≈ 83.6 °C
Therefore, the final temperature of the coffee will be approximately 83.6 °C.