In a container of negligible mass, 020 kg of ice at an initial temperature of - 40.0 oC is mixed with a mass m of water that has an initial temperature of 80.0 oC. No heat is lost to the surroundings. If the final temperature of the system is 20.0 oC, what is the mass m of the water that was initially at 80.0 oC

The mass is  

Explanation:

From the question we are told that    

     The mass of ice is  

     The  initial temperature of the ice is  

     The  initial temperature of the water is  

     The  final temperature of the system is

Generally according to the law of energy conservation,

   The  total heat loss is  =  total heat gained

 Now the total heat gain is mathematically represented as

Here   is the energy required to move the ice from

And it mathematically evaluated as

Here the specific heat of ice is  

So  

is the energy to melt the ice

And it mathematically evaluated as

The  latent heat of fusion of ice is  

So  

is the energy to raise the melted ice to

And it mathematically evaluated as

Here the specific heat of water  is  

So

The  heat loss is mathematically evaluated as

So

=>

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft