Finishing his ginger ale, Ramesh stands at the party holding his insulated foam cup that has nothing in it but 0.100kg of ice at 0 degree celsius. How much heat must be gained by the ice in order for all of it to melt? How much more heat must be gained to raise the temperature of the melted ice to room temperature of 23.0 celsius?
First question I have it: 3.35 x 10^4J
but second I don't know how to do it. Please, could you help me??? Thank you so much
How did you get the first answer?
Your answer to the first question is correct. For part 2, you need to used the specific heat C, the mass M, and the temperature rise (delta T = 23 C)
The specific heat of water is
C = 4.184 J/(deg*gram)
Q = M*C*(delta T) = 100*4.184*23 = ___ J
9623.2 J
Well, Ramesh seems to have a cool drink in his hand! Let's dive into the physics of it.
To determine how much heat must be gained by the ice in order for it to completely melt, we can use the equation:
Q = m * L
Where Q is the heat, m is the mass, and L is the latent heat of fusion of the substance (in this case, ice).
For ice, the latent heat of fusion is approximately 334,000 J/kg.
So, for the first question, where the mass of the ice is 0.100 kg, we can calculate the heat as:
Q = 0.100 kg * 334,000 J/kg
Q = 33,400 J
Thus, approximately 33,400 J of heat must be gained by the ice for it to completely melt.
Now, for the second part of your question, we need to determine how much more heat is required to raise the temperature of the melted ice to room temperature (23.0 °C).
To calculate this, we can use the equation:
Q = m * c * ΔT
Where Q is the heat, m is the mass, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is about 4,186 J/kg·°C.
Now, since the ice has already melted, we have water, not ice. So, we need to find the mass of the melted ice. Since 0.1 kg of ice melted, it will have the same mass in liquid water form.
So, we can calculate the heat as:
Q = 0.100 kg * 4,186 J/kg·°C * (23.0 °C - 0 °C)
Q ≈ 9,640 J
Therefore, approximately 9,640 J of heat must be gained to raise the temperature of the melted ice to room temperature.
I hope that helped! And remember, don't let the physics "cool" you down!
To solve this problem, we need to calculate the amount of heat required to melt the ice and then the amount of heat required to raise the temperature of the melted ice to room temperature.
To answer the first question, we need to determine the heat required to melt all the ice. The heat required to melt a substance can be calculated using the formula:
Q = m * L
where Q is the heat required, m is the mass of the ice, and L is the heat of fusion for ice.
The heat of fusion for ice is the amount of heat required to convert one kilogram of ice at 0 degrees Celsius to water at 0 degrees Celsius, which is 333,500 J/kg.
In this case, the mass of the ice is 0.100 kg, so we can substitute the values into the formula:
Q = 0.100 kg * 333,500 J/kg = 33,350 J
Therefore, the amount of heat required to melt all the ice is 33,350 J (or approximately 3.35 x 10^4 J), which answers your first question.
Now, to answer your second question, we need to calculate the amount of heat required to raise the temperature of the melted ice from 0 degrees Celsius to room temperature, which is 23.0 degrees Celsius.
The heat required to raise the temperature of a substance is given by the formula:
Q = m * c * ΔT
Where Q is the heat required, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
For water, the specific heat capacity is approximately 4,184 J/(kg·°C).
In this case, since the ice has already melted, we need to consider only the mass of the melted ice, which is still 0.100 kg. The change in temperature is 23.0 °C - 0 °C = 23.0 °C.
Substituting the values into the formula:
Q = 0.100 kg * 4,184 J/(kg·°C) * 23.0 °C = 9,646.40 J
Therefore, the amount of heat required to raise the temperature of the melted ice to room temperature is approximately 9,646.40 J.