When ice at 0C melts to liquid water at 0C, it absorbs 0.334 kj of heat per gram. Suppose the heat needed to melt 31.5 g of ice is absorbed from the water contained in a glass. If this water has a mass of .210 kg and a temperature of 21.0C, what is the final temperature of the water? (Note that you will also have 31.5 g of water at 0C from the ice.)
My Homework
q=ms∆T
q=.210kg x 0.334 kJ/g x (final temp-21C)
This is where I get stuck I feel like I am forgetting a step (or maybe I am completely doing this wrong)but cannot find anything in my notes or text that helps. Can anyone help?
that did not help at all, in order for that expression to equal zero, the deta T or "(Tf-Ti)" term has to equal zero. Im stuck where you are Capacino, idk what im doing wrong, this was a test question on my last exam and had me stumped.
The sum of the heats gained is zero.
Assuming the ice starts at zero (Ice is usually colder).
31.3*334+210*4.18*(Tf-21)=0
solve for Tf
To solve this problem, you are on the right track with using the equation q = msΔT, where q represents the heat absorbed or released, m represents the mass, s represents the specific heat capacity, and ΔT represents the change in temperature. However, in this case, you have to consider two stages: first, the heat absorbed to melt the ice, and second, the heat absorbed to raise the temperature of the resulting water.
Let's break down the steps:
1. Calculating the heat absorbed to melt the ice:
The heat absorbed to melt the ice can be calculated using the equation:
q1 = msΔT1
where m = 31.5 g, s = 0.334 kJ/g°C, and ΔT1 = 0°C - (-0°C) = 0°C.
Substituting these values into the equation, we have:
q1 = (31.5 g)(0.334 kJ/g°C)(0°C - (-0°C))
q1 = 0 kJ
This means that no heat is absorbed to raise the temperature of the ice to its melting point since it is already at 0°C.
2. Calculating the heat absorbed to raise the temperature of the water:
The heat absorbed to raise the temperature of the water can be calculated using the equation:
q2 = msΔT2
where m = 0.210 kg, s = 4.18 kJ/kg°C (specific heat capacity of water), and ΔT2 is the change in temperature we need to find.
Substituting these values into the equation, we have:
q2 = (0.210 kg)(4.18 kJ/kg°C)(ΔT2 - 21°C)
3. Since the total heat absorbed, q, is the sum of the heat absorbed in step 1 and step 2:
q = q1 + q2
0 kJ + (0.210 kg)(4.18 kJ/kg°C)(ΔT2 - 21°C)
4. We know that the total heat absorbed, q, is equal to zero since it is a closed system. Thus, we can set up the equation:
0 = (0.210 kg)(4.18 kJ/kg°C)(ΔT2 - 21°C)
Now you can solve for ΔT2, which represents the final temperature of the water.
To solve this problem, you need to use the principle of conservation of energy.
First, let's calculate the heat required to melt the ice. We know that 31.5 grams of ice requires 0.334 kJ of heat per gram to melt. Therefore, the total heat required to melt 31.5 grams of ice can be calculated as:
Heat (q) = mass (m) x heat absorbed per gram (s)
q = 31.5 g x 0.334 kJ/g
Now, let's calculate the heat gained by the water to reach the final temperature. Assuming no heat is lost to the surroundings, the heat gained by the water can be calculated using the equation:
Heat (q) = mass (m) x specific heat capacity (c) x temperature change (∆T)
We are given that the mass of the water is 0.210 kg and the initial temperature is 21.0°C. The final temperature is what we need to determine.
Using the equations above, we can set up the equation:
0.210 kg x c x (∆T) = q
Since the ice and water are in thermal equilibrium after the ice melts, the heat gained by the water is equal to the heat lost by the ice:
0.210 kg x c x (∆T) = 31.5 g x 0.334 kJ/g
Now, substitute the known values into the equation and solve for the final temperature (∆T):
0.210 kg x c x (∆T) = (31.5 g x 0.334 kJ/g)
From here, you can rearrange the equation to solve for (∆T) and calculate the final temperature.