200 g of ic is at -5C is dropped into a container containing water at 65C. If the final temperature of the mixture reaches 21C, what mass of water must be present in the container?

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

To find the mass of water present in the container, we can use the principle of conservation of energy. The heat lost by the ice will be equal to the heat gained by the water in the container.

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.