Ice
m₁=200 g = 0.2 kg
c₁ = 2060 J/kg•K
λ =335000 J/kg
Water
c₂=4183 J/kg•K
m₂=?
m₁c₁[0-(-5)] + m₁λ+ m₁c₂(21-0) = m₂c₂(65-21)
m₂=m₁(5c₁+λ+21c₂)/44 c₂ =
=0.2(5•2060 +335000+21•4183)/44•4183 =
=0.2(103000+335000+87843)/184052 =
=0.57 kg
m₁=200 g = 0.2 kg
c₁ = 2060 J/kg•K
λ =335000 J/kg
Water
c₂=4183 J/kg•K
m₂=?
m₁c₁[0-(-5)] + m₁λ+ m₁c₂(21-0) = m₂c₂(65-21)
m₂=m₁(5c₁+λ+21c₂)/44 c₂ =
=0.2(5•2060 +335000+21•4183)/44•4183 =
=0.2(103000+335000+87843)/184052 =
=0.57 kg
First, let's calculate the heat lost by the ice using the formula:
Q = m * c * ΔT
Where:
Q - Heat transferred
m - Mass of the ice
c - Specific heat capacity of ice
ΔT - Change in temperature
The specific heat capacity of ice is 2.09 J/(g·°C).
Q (heat lost by the ice) = m (mass of the ice) * c (specific heat capacity of ice) * ΔT (change in temperature)
Q = 200 g * 2.09 J/(g·°C) * (-5°C - 21°C)
Q = -200 g * 2.09 J/(g·°C) * (-26°C)
Q = 10,360 J
Since the heat lost by the ice is gained by the water, we can say that the heat gained by the water is also 10,360 J.
Now, let's calculate the heat gained by the water:
Q = m * c * ΔT
Where:
Q - Heat transferred
m - Mass of water
c - Specific heat capacity of water
ΔT - Change in temperature
The specific heat capacity of water is 4.18 J/(g·°C).
10,360 J = m (mass of water) * 4.18 J/(g·°C) * (21°C - 65°C)
10,360 J = m * 4.18 J/(g·°C) * (-44°C)
10,360 J = -m * 183.92 J
m = -10,360 J / (-183.92 J/g)
m ≈ 56.35 g
Therefore, approximately 56.35 grams of water must be present in the container.