You place a bottle containing 2.6 Liters of mineral water at 25*C into a refrigerator to cool to 4*C. Remember to show work when necessary.
a. How many kJ of heat are lost by the water?
b. How many kJ of heat are absorbed by the refrigerator?
c. What assumptions did you make in your calculations?

The answer to your question is below

Explanation:

data

volume = 2.6 l

temperature 1 = 25°C

temperature 2 = 4°C

density = 1 g/ml

specific heat = 4.186 J/g

Process

1.- Convert volume to milliliters

         1000 ml 1 l

             x         2.6 l

             x = (2.6 x 1000)/1

             x = 2600 ml

2.- Calculate the mass of water

mass = density x volume

         = 1 x 2600

         = 2600 g

3.- Calculate the heat (Q)

Formula

           Q = mC(T2 - T1

-Substitution

            Q = (2600)(4.186)(4 - 25)

            Q = (2600)(4.186)(-21)

            Q = -228555.6 J  or - 228.6 kJ

a)     Heat lost = -228.6 kJ

b)     Heat absorbed = 228.6 kJ

c)     The assumptions I made were the specific heat of water, the density of water, lost of heat is negative and absorption of heat is positive.

the answer is barium and copper bromide