A 70 piece of ice at 0 is added to a sample of water at 6. All of the ice melts and the temperature of the water decreases to 0. How many grams of water were in the sample?
You need units on the numbers. 70 what? 6 what? etc.
Oh my gosh, sorry.
It's a 70g piece of ice at 0 degrees celsius is added to a sample of water at 6 degrees celsius. All of the ice melts and the temp of the water decreases to 0 degrees Celsius.
After searching other posts, I think I found the formula.
Mass x Heat of fusion + [mass water x specific heat of water x (tfinal - t initial)]=0 Is this correct?
I need to solve for mass of water, I just can't seem to wrap my head around how to do that. This seems like a foreign language to me. Thanks for your help. I have been working on this problem for 4 days. :(
To find the amount of water in the sample, we can use the principles of heat transfer and the specific heat capacity of water.
First, let's understand the process that occurs:
1. The 70 grams of ice, which is at 0 degrees Celsius, are added to the water sample, which is at 6 degrees Celsius.
2. The ice absorbs heat from the water sample and starts melting.
3. The temperature of the water decreases to 0 degrees Celsius as the ice melts completely.
To solve this problem, we need to consider the heat exchange that takes place during the process.
The heat lost by the water is equal to the heat gained by the ice when it melts.
The heat exchange can be calculated using the formula Q = m * c * ΔT, where:
- Q is the heat exchange
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature
For water, the specific heat capacity (c) is approximately 4.18 J/g°C.
We can calculate the heat lost by the water:
Q_water = m_water * c_water * ΔT_water
Since the water temperature changes from 6°C to 0°C, the change in temperature (ΔT_water) is -6°C.
Q_water = m_water * 4.18 J/g°C * (-6°C)
On the other hand, the heat gained by the ice is equal to its heat of fusion, which is the energy required to melt a substance:
Q_ice = m_ice * ΔH_fusion
The heat of fusion (ΔH_fusion) for ice is 334 J/g.
Since all the ice is melted, its change in temperature is 0, and the heat gained by the ice is equal to the heat lost by the water. Therefore:
Q_water = Q_ice
m_water * 4.18 J/g°C * (-6°C) = m_ice * 334 J/g
We know that the mass of the ice (m_ice) is 70 g.
70 g * 334 J/g = m_water * 4.18 J/g°C * (-6°C)
Now, we can solve for the mass of the water (m_water):
m_water = (70 g * 334 J/g) / (4.18 J/g°C * (-6°C))
m_water ≈ 1,984 g
Therefore, there were approximately 1,984 grams of water in the sample.