A 63 gram block of an unknown metal at 88 degrees C was dropped into an insulated flask containing approx. 30 g of ice and 30 g of water at 0 degrees C. After the system had reached a steady temp. it was determined that 12.1 g of ice had melted. What is the specific heat of the metal?
Ok so I know that this problem involves the specific heat equation: q= mass x constant x change in temp. And I know that the heat of fusion of ice is 333 J/g. But I'm not sure of how to rearrange this equation or where to plug in the numbers.
If all the ice did not melt, there was no change in temperature.
Heat change= massicemelted*Heatfusionice
I guess I still don't really understand...
Bob Pursley is saying that a mixture of ice and water at zero degrees C will stay at that temperature until all of the ice is melted. If you had ice left after the experiment was performed, the final T of the ice/water mixture must still be zero degrees C. Thus the unknown metal melted ice (at 334 J/g) but it didn't raise the T of the mixture. (Therefore, you don't have a delta T)
Ok I see. But he said that heat change = the mass of the ice melted x the heat of fusion. Once I come up with this number, what would the next step be?
Do I have to rearrange the specific heat equation?
Ahh n/m ok I got it. :)
What answer did you obtain?
To find the specific heat of the metal, we need to use the principle of conservation of energy. In this case, the heat lost by the metal is equal to the heat gained by the ice and water.
First, calculate the heat absorbed by the ice to melt it:
Heat absorbed by ice = mass of ice melted * heat of fusion of ice
= 12.1 g * 333 J/g
= 4029.3 J
Now, since the system has reached a steady temperature, we know that the heat lost by the metal is equal to the heat gained by the ice and water:
Heat lost by metal = Heat gained by ice + Heat gained by water
The heat gained by the ice is the heat absorbed by the ice calculated above.
The heat gained by the water is given by the equation:
Heat gained by water = mass of water * specific heat of water * change in temperature
As the water is initially at 0 degrees C and reaches a steady temperature, there is no change in temperature. Hence, the heat gained by the water is zero.
Therefore, we have:
Heat lost by metal = Heat gained by ice + Heat gained by water
Heat lost by metal = 4029.3 J + 0 J
Heat lost by metal = 4029.3 J
Now, we can use the specific heat equation:
Heat lost by metal = mass of metal * specific heat of metal * change in temperature
The mass of the metal is given as 63 g and the change in temperature is from its initial temperature of 88 degrees C to the final temperature.
Since the final temperature is not given, we assume that the final temperature of the mixture of ice and water is 0 degrees C. Therefore, the change in temperature is 88 degrees C.
Now, rearrange the equation and solve for the specific heat of the metal:
specific heat of metal = Heat lost by metal / (mass of metal * change in temperature)
= 4029.3 J / (63 g * 88 degrees C)
≈ 0.820 J/(g°C)
Therefore, the specific heat of the metal is approximately 0.820 J/(g°C).