Suppose four 30-g ice cubes are added to a glass containing 200 cm3 of orange juice at

Question

Suppose four 30-g ice cubes are added to a glass containing 200 cm3 of orange juice at

20,0°C.
You may assume that the orange juice has a density and specific heat capacity the same
as water namely: Density = 1,0 g∙cm–3 and specific heat = 4,184 J/gK
If your finding is that some ice will be left in the juice, you must calculate what mass
of ice will be present the moment when thermal equilibrium is reached.

Answer 1

First, we need to calculate the initial temperature of the orange juice before adding the ice cubes. Since the orange juice is at 20.0°C, we can calculate the initial heat content using the formula:

Q = mcΔT

Where:
Q = heat content
m = mass of orange juice
c = specific heat capacity of water
ΔT = change in temperature

Given:
m = 200 cm3 = 200 g (since density = 1.0 g∙cm–3)
c = 4.184 J/gK
ΔT = (20.0°C - 0°C) = 20.0 K

So, Q = 200 g x 4.184 J/gK x 20.0 K = 1673.6 J

Now, let's calculate the amount of heat required to melt the ice cubes and raise the temperature of the resulting water to the final equilibrium temperature. The heat gained by the ice cubes will be equal to the heat lost by the orange juice:

Q_lost = Q_gained
m_orange_juice x c x ΔT = m_ice x L_f + m_ice x c x ΔT

Where:
m_orange_juice = 200 g
c = specific heat capacity of water = 4.184 J/gK
ΔT = final equilibrium temperature - initial temperature = 0°C - 20.0°C = -20.0 K
L_f = latent heat of fusion of ice = 334 J/g

Substitute the values into the equation:

200 g x 4.184 J/gK x -20.0 K = m_ice x 334 J/g + m_ice x 4.184 J/gK x -20.0 K

Solve for m_ice:

-83680 J = 334 m_ice + (-83.68 m_ice)
-83680 J = -250.32 m_ice
m_ice = -83680 J / -250.32
m_ice ≈ 334 g

Therefore, approximately 334 g of ice will be present when thermal equilibrium is reached.