I was having trouble with two different physics problems when studying for a test. Any help is appreciated.
1.) A 11.6-kg block of ice has a temperature of -14.5 °C. The pressure is one atmosphere. The block absorbs 6.87 x 106 J of heat. What is the final temperature in degrees Celsius of the liquid water?
2.) A thermos contains 156 cm3 of coffee at 86.4 °C. To cool the coffee, you drop two 11.4-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.
Thanks
Sure, I'd be happy to help you with these physics problems and also explain how to solve them.
1.) To find the final temperature of the liquid water, you can use the specific heat formula:
Q = mcΔT
where Q is the heat absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.
In this problem, you are given:
- Mass of the block (m) = 11.6 kg
- Heat absorbed (Q) = 6.87 x 10^6 J (or 6,870,000 J)
- Initial temperature of the block = -14.5 °C
First, you need to determine the heat absorbed per kilogram of the block. Divide the total heat absorbed by the mass of the block:
Q/m = (6.87 x 10^6 J) / (11.6 kg) ≈ 592,241 J/kg
Next, you need to use the specific heat of water (c) to find the change in temperature (ΔT). The specific heat of water is approximately 4190 J/kg°C.
Using the formula Q = mcΔT, you can rearrange it to solve for ΔT:
ΔT = Q / (mc)
Substituting the known values:
ΔT ≈ (592,241 J/kg) / (11.6 kg * 4190 J/kg°C) ≈ 12.9 °C
Lastly, to find the final temperature, you need to add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature ≈ -14.5 °C + 12.9 °C ≈ -1.6 °C
Therefore, the final temperature of the liquid water is approximately -1.6 °C.
2.) To find the final temperature of the coffee, you can use the principle of conservation of energy. The heat lost by the coffee will be equal to the heat gained by the ice cubes, given that the coffee is treated as water.
First, you need to find the heat lost by the coffee. You can use the specific heat formula:
Q = mcΔT
In this problem, you are given:
- Volume of the coffee (V) = 156 cm^3
- Initial temperature of the coffee = 86.4 °C
- Density of water (ρ) = 1 g/cm^3 (since we are treating the coffee as water)
- Specific heat of water (c) = 4.186 J/g°C
First, you need to convert the volume of coffee to mass:
First, we need to convert the volume of coffee to mass. Since density (ρ) = mass (m) / volume (V), we can rearrange it to solve for mass:
m = ρ * V
Substituting the known values:
m ≈ (1 g/cm^3) * (156 cm^3) ≈ 156 g
Next, you can use the specific heat formula to find the heat lost by the coffee:
Q = mcΔT
Substituting the known values and rearranging the formula to solve for ΔT:
ΔT = Q / (mc)
Since the heat gained by the ice cubes is equal to the heat lost by the coffee, the heat gained by the ice cubes can be calculated by:
Q = mcΔT
In this case, you are given:
- Mass of the ice cubes (m) = 11.4 g
- Initial temperature of the ice cubes (which is also 0 °C)
- ΔT of the ice cubes (since they melt completely and heat up to 0 °C)
Again, use the specific heat formula to calculate ΔT:
ΔT = Q / (mc)
Since the heat gained by the ice cubes is equal to the heat lost by the coffee, you can set the two equations equal to each other:
mcΔT(of coffee) = mcΔT(of ice)
Substituting the known values, you can solve for ΔT(of coffee):
156 g * 4.186 J/g°C * (ΔT(of coffee)) = 11.4 g * 4.186 J/g°C * (0 - 0 °C)
Simplifying the equation, you will get:
ΔT(of coffee) = (11.4 g * 4.186 J/g°C * 0) / (156 g * 4.186 J/g°C)
Therefore, ΔT(of coffee) = 0
Finally, to find the final temperature of the coffee, you need to add the change in temperature (which is 0 in this case) to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 86.4 °C + 0 °C = 86.4 °C
So, the final temperature of the coffee is 86.4 °C.
I hope this explanation helps! Let me know if you have any further questions.