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We have a glass containing 0.5 liter (0.5 kg) of water whose temperature 100 degrees Celsius.
We also have an ice cube with mass 0.01 kg and temperature -10 degrees Celsius.
The cube is put into the glass. The glass is then insulated from the outside world, until the cube has melted. What will be the terminal temperature of the water?
I have read a little bit about Thermodynamics and I think this solution shoudl work:
Let's first calculate the amount of energy required to melt the cube. First, we need to raise its temperature by 10 degrees. The specific heat of ice is 2108 J /(kgK) and we have 0.01 kg of it, and so we will need
$$E_1 = 0.01 \cdot 2108 \cdot 10 = 210.8 J$$
Now, the ice needs to transition to the liquid state.
The amount of heat required to melt one kilogram of ice is 333 700 J. Thus, we need
$$E_2 = 0.01 \cdot 333 700 = 3337 J$$
So, we need the total of
$$\Delta E = 3547.8 J$$
And this energy will be sucked away from the water. So now we can sovle for the change in temperature
$$3547.8 = 0.5 \cdot 4180 \cdot \Delta T$$
$$\Delta T = 1.7 $$ Degrees Celsius.
Is it the correct answer? It seems to me that the water will be way too hot.
We also have an ice cube with mass 0.01 kg and temperature -10 degrees Celsius.
The cube is put into the glass. The glass is then insulated from the outside world, until the cube has melted. What will be the terminal temperature of the water?
I have read a little bit about Thermodynamics and I think this solution shoudl work:
Let's first calculate the amount of energy required to melt the cube. First, we need to raise its temperature by 10 degrees. The specific heat of ice is 2108 J /(kgK) and we have 0.01 kg of it, and so we will need
$$E_1 = 0.01 \cdot 2108 \cdot 10 = 210.8 J$$
Now, the ice needs to transition to the liquid state.
The amount of heat required to melt one kilogram of ice is 333 700 J. Thus, we need
$$E_2 = 0.01 \cdot 333 700 = 3337 J$$
So, we need the total of
$$\Delta E = 3547.8 J$$
And this energy will be sucked away from the water. So now we can sovle for the change in temperature
$$3547.8 = 0.5 \cdot 4180 \cdot \Delta T$$
$$\Delta T = 1.7 $$ Degrees Celsius.
Is it the correct answer? It seems to me that the water will be way too hot.