An insulated beaker with negligible mass contains a mass of 0.300 kgkg of water at a temperature of 71.5°C. Take the specific heat for water to be 4190 J/kg⋅K, the specific heat for ice to be 2100 J/kg⋅K, and the heat of fusion for water to be 334 kJ/kgJ. Required:
How much ice at a temperature of -20.9 ∘C must be dropped into the water so that the final temperature of the system will be 40.0°C ?

The mass of ice to be added is  

Explanation:

From the question we are told that

      The  mass of water is  

      The temperature of water   is  

      The  specific heat of water is  

       The specific heat of ice is  

         The heat of fusion of water is  

        The final temperature is  

         The temperature of ice is  

Generally according to the law of  energy conservation

   The  heat gained by ice  =  The  heat lost by water

 The  heat lost by water is mathematically represented as

substituting values

The  amount of heat gained by the ice is

Where   is the amount of heat absorbed by the ice to get to   (273 K)and this  is mathematically represented as

Here  M is the mass of ice

  substituting values

  is the heat gained by the ice as it is been converted to liquid

  substituting values

And   is the heat gained by the ice(now water) as it is been heated to and it is mathematically represented as

  substituting values

So  

From the law of energy conservation

=>    

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.