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 can help you with both of these physics problems.
1.) To solve this problem, we need to use the specific heat formula:
Q = m * c * ΔT
where Q is the heat absorbed or released by the substance, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
In this case, we have the mass of the ice block (m = 11.6 kg), the heat absorbed by the ice block (Q = 6.87 x 10^6 J), and the initial temperature of the ice block (-14.5 °C).
Now, we need to find the specific heat capacity of ice. The specific heat capacity of ice is 2.09 J/g°C, but we need to convert the units to match the mass given in kilograms:
Specific heat capacity of ice = 2.09 J/g°C = 2090 J/kg°C
Now, let's plug the values into the formula:
6.87 x 10^6 J = (11.6 kg) * (2090 J/kg°C) * ΔT
Solving for ΔT:
ΔT = (6.87 x 10^6 J) / (11.6 kg * 2090 J/kg°C) = 290.12 °C
Finally, we add the change in temperature to the initial temperature to find the final temperature:
Final temperature = -14.5 °C + 290.12 °C = 275.62 °C
Therefore, the final temperature of the liquid water will be approximately 275.62 °C.
2.) In this problem, we need to use the principle of conservation of energy to find the final temperature of the coffee.
First, let's find the heat absorbed by the coffee. We can use the formula:
Q = m * c * ΔT
where Q is the heat absorbed by the substance, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
In this case, we have the mass of the coffee (156 cm^3) and the initial and final temperatures of the coffee.
To find the mass of the coffee, we need to convert the volume from cubic centimeters to grams:
156 cm^3 = 156 g
The specific heat capacity of water is 4.18 J/g°C. Now, let's plug the values into the formula:
Q = (156 g) * (4.18 J/g°C) * ΔT
The heat absorbed by the coffee is equal to the heat released by the melted ice cubes. Since the ice cubes are initially at 0 °C, we can calculate the heat released using the formula:
Q = (11.4 g) * (334 J/g) = 3810.6 J
Now, equating the two equations, we get:
(4.18 J/g°C) * ΔT = 3810.6 J
Solving for ΔT:
ΔT = 3810.6 J / 4.18 J/g°C = 913.93 °C
Finally, we add the change in temperature to the initial temperature of the coffee to find the final temperature:
Final temperature = 86.4 °C + 913.93 °C = 1000.33 °C
Therefore, the final temperature of the coffee will be approximately 1000.33 °C.
I hope this helps! Let me know if you need further clarification or assistance with any other questions.