What mass of water at 10.0 oc would be required to cool 50.0g of a metal having a specific heat of
0.60 j/g. oc from 90.0 oc to 20.0 oc?

The mass of water required is 50.2 g

That is;

Q=m×c×ΔT

In this case, water is used to cool a metal, therefore, water will gain heat while the metal will lose heat.

Heat gained is equivalent to heat lost

Heat gained by water = Heat lost by the metal

Step 1: Heat lost by the metal

Mass of the metal  = 50.0 g

Specific heat capacity of metal = 0.60 J/g°C

Change in temperature (ΔT) = 70°C

Heat lost by the metal = 50 g× 0.6 × 70

                                     = 2100 Joules

Step 2: Heat gained by water

Mass of water = x g

Specific heat capacity of water = 4.18 J/g°C

Temperature change = 10 °C

Heat gained by water = x g × 4.18 × 10

                                    = 41.8x joules

Step 3: Mass of water

Heat gained by water = heat lost by the metal

41.8x = 2100

   x = 50.239 g

      = 50.2 g (1 d.p)

The mass of water is 50.2 g