1. A 0.400 kg sample of aluminum (c = 9.20 x 102 J/kg°C) at 95.0 °C is dropped into a 0.550 kg pot of water which is at 18.0 °C. What temperature will the mixture come to?

2. How much thermal energy is absorbed by a 0.0220 kg ice cube as it melts? The latent heat of fusion of water is 3.40x 10^5 J/kg.

To solve these problems, we need to apply the concepts of specific heat and latent heat of fusion. Here's how we can find the answers:

1. To solve for the final temperature of the mixture, we can use the principle of heat transfer. The equation is given by:

m1c1(T1 - Tf) + m2c2(T2 - Tf) = 0

Where:
m1 = mass of aluminum
c1 = specific heat of aluminum
T1 = initial temperature of aluminum
Tf = final temperature of the mixture
m2 = mass of water
c2 = specific heat of water
T2 = initial temperature of water

Given data:
m1 = 0.400 kg
c1 = 9.20 x 10^2 J/kg°C
T1 = 95.0 °C
m2 = 0.550 kg
c2 = specific heat of water (we'll assume 4.18 x 10^3 J/kg°C, which is the value for liquid water)
T2 = 18.0 °C

Substituting the given values into the equation, we get:

(0.400 kg)(9.20 x 10^2 J/kg°C)(95.0 °C - Tf) + (0.550 kg)(4.18 x 10^3 J/kg°C)(18.0 °C - Tf) = 0

Now, we can solve for Tf using algebraic calculations:

(0.400 kg)(9.20 x 10^2 J/kg°C)(95.0 °C) + (0.550 kg)(4.18 x 10^3 J/kg°C)(18.0 °C) = (0.400 kg)(9.20 x 10^2 J/kg°C)(Tf) + (0.550 kg)(4.18 x 10^3 J/kg°C)(Tf)

Simplifying further:

35,040 J + 41,166 J = (0.400 kg)(9.20 x 10^2 J/kg°C + 0.550 kg)(4.18 x 10^3 J/kg°C) Tf

76,206 J = (3.68 x 10^3 J/kg°C + 2.299 J/kg°C) Tf

76,206 J = 5.979 x 10^3 J/kg°C Tf

Tf = 76,206 J / 5.979 x 10^3 J/kg°C
Tf ≈ 12.72 °C

Therefore, the final temperature of the mixture is approximately 12.72 °C.

2. To find the thermal energy absorbed by the ice cube as it melts, we need to use the equation:

Q = m ΔH

Where:
Q = thermal energy absorbed
m = mass of the ice cube
ΔH = latent heat of fusion of water

Given data:
m = 0.0220 kg
ΔH = 3.40 x 10^5 J/kg

Substituting the given values into the equation:

Q = (0.0220 kg)(3.40 x 10^5 J/kg)
Q ≈ 7,480 J

Therefore, the thermal energy absorbed by the ice cube as it melts is approximately 7,480 J.