The mass of ice needed to cool 60 g of water from 43 °C to 20 °C is _____? [Take the specific heat capacity of water as 4200 Jkg-1 °C-1 and specific latent heat of fusion of ice as 3.3×105 J/kg]
To calculate the mass of ice needed to cool the water, we can break the process into two parts: cooling the water from 43 °C to 0 °C and then freezing the water at 0 °C to ice at 0 °C.
1. Cooling the water from 43 °C to 0 °C:
The heat lost by water can be calculated using the equation:
Q = mcΔT
Where:
Q = heat lost
m = mass of water (60 g)
c = specific heat capacity of water (4200 J/kg °C)
ΔT = change in temperature (final temperature - initial temperature)
Q = 60 g * 4200 J/kg °C * (0 °C - 43 °C)
Q = 60 g * 4200 J/kg °C * (-43 °C)
Q = - 109,200 J
2. Freezing the water at 0 °C to ice at 0 °C:
The heat lost during the phase change from water to ice can be calculated using the equation:
Q = mL
Where:
Q = heat lost
m = mass of water turned into ice (unknown)
L = specific latent heat of fusion of ice (3.3×105 J/kg)
Q = m * 3.3×105 J/kg
Q = - 109,200 J
Since the heat lost during the cooling process equals the heat lost during the phase change, we can equate the two equations:
-109,200 J = m * 3.3×105 J/kg
Solving for m:
m = (-109,200 J) / (3.3×105 J/kg)
m ≈ - 0.33 kg
The mass of ice needed to cool the water from 43 °C to 20 °C is approximately 0.33 kg.