Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: ice = 2.06 J/g·K, liquid water = 4.184 J/g·K)

Explanation:

The given data is as follows.

          mass, m = 75 g

      Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

            q =

               =

               = 8464.5 J/mol

               = 8.46 kJ ......... (1)

Also, it is given that = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

        q =

           =

           = -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

               8.46 kJ - 46.32 kJ

             = -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at to ice at is -37.86 kJ.

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.

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