Q.an aluminium can of mass 200gm contains 500gms of water at 50c.an ice cube of mass 50gms,initially at a temperature of -20c is thrown into the can of water.
i.what is the final temperature onces the ice metls and comes into equilibrium?
ii.now a 200gm piece of Iron whose temperature is 200c is thrown into the can.what is the new equilibrium temperature.
specific heats:cal/gm-deg
water-1.0
ice-0.5
Iron-0.1
Al-0.2
finally what mass of ice would have to be thrown in initially so that it just barely melts?
To solve the given questions, we will use the principle of conservation of energy.
i. To find the final temperature once the ice melts and comes into equilibrium:
Step 1: Calculate the heat lost by the aluminum can to cool down:
Heat lost by aluminum can = mass of aluminum can * specific heat of aluminum * change in temperature
Heat lost by aluminum can = 200g * 0.2 cal/gm-deg * (final temperature - 50 deg)
Step 2: Calculate the heat gained by the water:
Heat gained by water = mass of water * specific heat of water * change in temperature
Heat gained by water = 500g * 1.0 cal/gm-deg * (final temperature - 50 deg)
Step 3: Calculate the heat gained by the ice:
Heat gained by ice = mass of ice * specific heat of ice * change in temperature
Heat gained by ice = 50g * 0.5 cal/gm-deg * (final temperature - (-20 deg))
Since the ice melts at 0 degrees Celsius and reaches equilibrium, the heat gained by the ice will also be equal to the heat lost by the aluminum can and the heat gained by the water.
Hence, equating the three heat values:
Heat lost by aluminum can = Heat gained by water + Heat gained by ice
200g * 0.2 cal/gm-deg * (final temperature - 50 deg) = 500g * 1.0 cal/gm-deg * (final temperature - 50 deg) + 50g * 0.5 cal/gm-deg * (final temperature - (-20 deg))
Solving the equation will give the final temperature when the ice melts and comes into equilibrium.
ii. To find the new equilibrium temperature after a 200g piece of iron at 200 degrees Celsius is thrown into the can:
Step 1: Calculate the heat lost by the iron:
Heat lost by iron = mass of iron * specific heat of iron * change in temperature
Heat lost by iron = 200g * 0.1 cal/gm-deg * (final temperature - 200 deg)
Step 2: Calculate the new equilibrium temperature by equating the heat lost by the iron to the sum of heat gained by the water and the heat gained by the aluminum can:
Heat lost by iron = Heat gained by water + Heat gained by aluminum can
200g * 0.1 cal/gm-deg * (final temperature - 200 deg) = 500g * 1.0 cal/gm-deg * (final temperature - final temperature found in (i)) + 200g * 0.2 cal/gm-deg * (final temperature - final temperature found in (i))
Solving the equation will give the new equilibrium temperature.
iii. To find the mass of ice to be thrown in initially so that it just barely melts:
We know that the heat gained by the ice will be equal to the heat lost by the aluminum can and the heat gained by the water.
Mass of ice * specific heat of ice * (final temperature - (-20 deg)) = 200g * 0.2 cal/gm-deg * (final temperature - 50 deg) + 500g * 1.0 cal/gm-deg * (final temperature - 50 deg)
Solving the equation will give the mass of ice required to just barely melt.
Please note that the specific heat values are given in cal/gm-deg and the final temperatures will be in degrees Celsius.
To find the final temperature and the equilibrium temperature in the given scenarios, we'll use the principle of conservation of energy. The heat gained by one object is equal to the heat lost by the other object.
i. Let's consider the first scenario where the ice cube is thrown into the can of water.
1. Calculate the heat gained by the water:
Heat gained by water = mass of water * specific heat of water * change in temperature
= 500g * 1.0 cal/gm-degC * (final temperature - 50°C)
2. Calculate the heat lost by the ice cube:
Heat lost by ice = mass of ice * specific heat of ice * change in temperature
= 50g * 0.5 cal/gm-degC * (0°C - (-20°C))
Since the ice melts and comes into equilibrium with the water, the heat gained by the water is equal to the heat lost by the ice cube. Therefore, we set the two equations equal to each other and solve for the final temperature.
Heat gained by water = Heat lost by ice
500 * (final temperature - 50) = 50 * (0 - (-20))
Solving the above equation will give us the final temperature.
ii. For the second scenario, where the piece of iron is thrown into the can of water, we follow a similar process:
1. Calculate the heat gained by the water:
Heat gained by water = mass of water * specific heat of water * change in temperature
= 500g * 1.0 cal/gm-degC * (final temperature - initial temperature)
2. Calculate the heat lost by the iron piece:
Heat lost by iron = mass of iron * specific heat of iron * change in temperature
= 200g * 0.1 cal/gm-degC * (200°C - initial temperature)
Again, since the iron and water reach equilibrium, we set the two equations equal to each other and solve for the equilibrium temperature.
Heat gained by water = Heat lost by iron
500 * (final temperature - initial temperature) = 200 * (200 - initial temperature)
Solving the above equation will give us the equilibrium temperature.
Finally, to determine the mass of ice that would just barely melt, we need to find the point where the heat gained by the ice equals the heat lost during the process. This can be done by setting the equations for heat gained and lost equal to each other.
I hope this helps!