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
