Suppose that 90 g of hot water at a temperature of 75 degrees C is poured into a cavity in a very large block of ice at 0 degrees C.

(a) What will be the final temperature of the water in the cavity?
(b) What mass of ice must melt in order for the water to cool down to the temperature found in Part (a)?

To solve this problem, we need to use the principles of heat transfer and energy conservation.

(a) To determine the final temperature of the water, we can use the principle of energy conservation, which states that the total energy of a closed system remains constant.

First, let's calculate the amount of heat gained or lost by the water using the formula:

Q = mcΔT

Where:
Q is the heat gained or lost
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

Given:
Initial mass of hot water, m1 = 90 g
Initial temperature of hot water, T1 = 75°C
Final temperature, T2 (to be determined)
Specific heat capacity of water, c1 = 4.18 J/g°C

Since the water is cooling down, the heat lost by the water equals the heat gained by the ice, so:

Q_lost = Q_gained
m1c1(T1 - T2) = m2c2(T2 - T0)

Where:
m2 is the mass of ice melted (to be determined)
c2 is the specific heat capacity of ice = 2.09 J/g°C
T0 is the melting point of ice = 0°C

Substituting the given values, we can solve for T2:

90 g * 4.18 J/g°C * (75°C - T2) = m2 * 2.09 J/g°C * (T2 - 0°C)

Simplifying the equation, we get:

376.2 g°C - 4.18 g * T2 = 2.09 m2 * T2

Rearranging the equation, we have:

4.18 g * T2 + 2.09 m2 * T2 = 376.2 g°C

Combining like terms, we get:

T2(4.18 g + 2.09 m2) = 376.2 g°C

Dividing both sides of the equation by (4.18 g + 2.09 m2), we have:

T2 = 376.2 g°C / (4.18 g + 2.09 m2)

This equation gives us the final temperature of the water in the cavity.

(b) To calculate the mass of ice melted, we can use the equation obtained earlier:

m2 = (90 g * 4.18 J/g°C - 375 g°C) / (2.09 J/g°C)

Simplifying the equation, we get:

m2 = (90 g * 4.18 J/g°C - 375 g°C) / 2.09 J/g°C

m2 = 378.27 g

Therefore, 378.27 grams of ice must melt in order for the water to cool down to the final temperature found in Part (a).