a): Two 50 g ice cubes are dropped into 200 g of water in a thermally insulated container. If the water is initially at 25 degree Celsius, and the ice comes directly from a freezer at -15 degree Celsius, what is the final temperature at thermal equilibrium? b): What is the final temperature if only one ice cube is used?

To answer both questions, we need to use the principles of heat transfer and the concept of thermal equilibrium.

a) Two 50 g ice cubes are dropped into 200 g of water:
When the ice cubes come into contact with the water, heat will transfer from the water to the ice until they reach thermal equilibrium, which means they will have the same final temperature.

To find this temperature, we need to consider the heat gained and lost by each component of the system.
- The heat gained by the ice cubes is given by the equation: Q = m * c * (Tf - Ti), where Q is the heat gained, m is the mass of the substance, c is the specific heat capacity, Tf is the final temperature, and Ti is the initial temperature.
- The specific heat capacity of ice is about 2.09 J/g°C, and the initial temperature of the ice is -15°C.
- The heat lost by the water can be calculated using the same equation: Q = m * c * (Tf - Ti), where m is the mass of water, c is the specific heat capacity of water (which is about 4.18 J/g°C), and the initial temperature of the water is 25°C.

Since the system is thermally insulated, the heat lost by the water is equal to the heat gained by the ice cubes. Therefore, we can set up an equation to solve for the final temperature, Tf:

(m1 * c1 * (Tf - Ti1)) + (m2 * c2 * (Tf - Ti2)) = 0

Substituting the given values:
(50g * 2.09 J/g°C * (Tf - (-15°C))) + (50g * 2.09 J/g°C * (Tf - (-15°C))) + (200g * 4.18 J/g°C * (Tf - 25°C)) = 0

Simplifying the equation will give us the final temperature (Tf) at thermal equilibrium.

b) If only one ice cube is used:
The same principles apply in this scenario. The heat gained by the ice cube and the heat lost by the water will still be equal at thermal equilibrium. However, since there is only one ice cube, the mass of ice (m2) will be zero in the equation.

Simplifying the equation with one ice cube will give us the final temperature (Tf) in this case.

By solving these equations, we will be able to find the final temperature at thermal equilibrium for both scenarios.

a) To solve this problem, we can use the principle of conservation of energy. The energy lost by the water will be equal to the energy gained by the ice cubes.

The heat gained by the ice cubes can be calculated using the formula: Q = m * c * ΔT, where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat lost by the water can be calculated using the same formula, considering the final temperature of both the water and ice as the temperature at thermal equilibrium.

Since the water and ice are in thermal equilibrium, the final temperature will be the same for both.

Now let's calculate the heat gained and lost:

For the ice cubes:
Q_ice = m_ice * c_ice * ΔT_ice

Given:
m_ice = 50 g (mass of one ice cube)
c_ice = 2.09 J/g°C (specific heat capacity of ice)
ΔT_ice = (-15°C) - (0°C) = -15°C (change in temperature)

Q_ice = 50 g * 2.09 J/g°C * (-15°C) = -1567.5 J

For the water:
Q_water = m_water * c_water * ΔT_water

Given:
m_water = 200 g (mass of water)
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = final temperature - initial temperature = final temperature - 25°C

Q_water = 200 g * 4.18 J/g°C * (final temperature - 25°C)

Since the total energy lost by the water is equal to the total energy gained by the ice cubes, we can write the following equation:

-1567.5 J = 200 g * 4.18 J/g°C * (final temperature - 25°C)

Simplifying the equation:
-1567.5 J = 836 J/°C * (final temperature - 25°C)

Dividing both sides by 836 J/°C:
-1567.5 J / 836 J/°C = final temperature - 25°C

-1.875 °C = final temperature - 25°C

Adding 25°C to both sides:
final temperature = -1.875 °C + 25°C = 23.125 °C

Therefore, the final temperature at thermal equilibrium is approximately 23.125 °C.

b) If only one ice cube is used, the mass of the ice cubes will be halved, resulting in a change in the calculation of the heat gained by the ice cubes. Let's calculate it:

For the ice cube:
m_ice = 25 g (half the mass of two ice cubes)

Q_ice = m_ice * c_ice * ΔT_ice
Q_ice = 25 g * 2.09 J/g°C * (-15°C) = -522.5 J

Using the same equation as before to find the final temperature, the equation becomes:

-522.5 J = 836 J/°C * (final temperature - 25°C)

Simplifying the equation:
-522.5 J = 836 J/°C * (final temperature - 25°C)

Dividing both sides by 836 J/°C:
-522.5 J / 836 J/°C = final temperature - 25°C

-0.624 °C = final temperature - 25°C

Adding 25°C to both sides:
final temperature = -0.624 °C + 25°C = 24.376 °C

Therefore, the final temperature at thermal equilibrium when only one ice cube is used is approximately 24.376 °C.

heat gained in anyting changing temp =mass*c*(tf-ti)

heat gained by melting ice: mass*Lf where Lf is the heat of fusion for ice per mass.

Now, when this happens in a closed system, everything has the same final temperature, and the sum of all heats gained is zero (some of them will be negative0.

Heat gained by ice/melted ice+heat gained by water=0

Ice:
massice*cice*(0-(-15)) + massice*Lf + massmelted ice*cwater*(Tf-0)
Waterinitial
masswater*cwater*(Tf-25)

set both of these equal, and solve for Tf.