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

Explanation: Quantity of heat of a substance is calculated by multiplying the mass of a substance by the specific heat capacity and the change in temperature.

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