A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108.7 g of water, already in the calorimeter, at 19.3°C, the final temperature of the water is 35.0°C. Calculate the heat capacity of the calorimeter in J/K. Use 4.184 J/g°C as the specific heat of water. Enter to 1 decimal place.

Answer : The heat capacity of the calorimeter is,

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

where,

= specific heat of water =

= specific heat of calorimeter = ?

= mass of water = 108.7 g

= mass of calorimeter = 108.7 g

= final temperature of mixture =

= initial temperature of water =

= initial temperature of calorimeter =

Now put all the given values in the above formula, we get

Therefore, the heat capacity of the calorimeter is,

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

i think it would be 5