An 850 gram chunk of ice at 0 degrees Celsius is dropped into a container holding 1.7 kg of water at an initial temperature of 35 degrees Celsius. Due to the presence of the ice, the temperature of the water eventually drops to 0 degrees Celsius. Show a mathematical solution to answer this question:
Does the 850 gram chunk of ice completely melt?
Formulas: Q= mHf Q=mHv Q= m∆TCp
The specific heat of liquid is 4.18J/gᵒC
The specific heat of solid water (ice) is 2.11J/gᵒC
The heat energy released during melting is 334J/g
heat to melt ice at zero C = 850 x heat fusion = ?
heat removed from 35 C water = 1,700 x specific heat water x (0-35) = ?
I get about 280,000 J to melt the ice and about 250,000 J from the water. The ice won't melt completely. The end solution will be a mixture of ice and water at zero C. You can calculate how much ice is left if you wish. There is enough data to do that.
Could you please prove how the ice doesn't completely melt? Don't you subtract 280,000 J and 250,000 J and get 30,000 J? and then you divide it by 334 J/g and get 105 g? But I don't know what you do next.
I caution you not to use my estimates. Those are just close numbers; you need to run them yourself.
What I did above SHOWS that the ice doesn't completely melt. Yes, you subtract 280,000 - 250,000 and show that you are 30,000 J short of having enough heat to melt all of the ice. I don't think anything else is necessary. Again, that 30,000 is an estimate. You need to run that number too.
To determine if the 850 gram chunk of ice completely melts, we need to calculate the amount of heat energy required to melt the ice and compare it to the amount of heat energy available in the water.
First, let's calculate the heat energy required to melt the ice using the formula Q = mHf, where Q is the heat energy, m is the mass, and Hf is the heat of fusion.
Given:
Mass of ice (m) = 850 grams
Heat of fusion (Hf) = 334 J/g
Q (heat energy required to melt the ice) = m * Hf
Q = 850 g * 334 J/g
Q = 284,900 J
The heat energy required to melt the ice is 284,900 J.
Next, let's calculate the heat energy available in the water using the formula Q = m∆TCp, where Q is the heat energy, m is the mass, ∆T is the change in temperature, and Cp is the specific heat capacity.
Given:
Mass of water (m) = 1.7 kg = 1700 grams
Initial temperature of the water (T1) = 35°C
Final temperature of the water (T2) = 0°C
Specific heat capacity of water (Cp) = 4.18 J/g°C
First, we need to calculate the change in temperature (∆T):
∆T = T2 - T1
∆T = 0°C - 35°C
∆T = -35°C
Now, we can calculate the heat energy available in the water:
Q (heat energy available in the water) = m * ∆T * Cp
Q = 1700 g * -35°C * 4.18 J/g°C
Q = -317,710 J
The heat energy available in the water is -317,710 J. Note that the negative sign indicates heat loss.
Comparing the heat energy required to melt the ice (284,900 J) and the heat energy available in the water (-317,710 J), we can conclude that the available heat energy in the water is not enough to completely melt the ice. Therefore, the 850 gram chunk of ice will not completely melt.