Question

A 18.5-g sample of ice at -10.0 °c is mixed with 110.0 g water at 85.0 °c. What will be the final temperature (in °c) of the mixture assuming no heat loss to the surroundings?

Answer 1

The final temperature of the mixture will be 19.22 °C.

Step-by-step explanation:

To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat gained by the ice will be equal to the heat lost by the water. The equation we can use to solve this problem is:

Qice + Qwater = 0

First, we calculate the heat gained by the ice:

Qice = mcΔT

where m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature. The specific heat capacity of ice is 2.09 J/g°C. Substitute the given values into the equation:

Qice = (18.5 g)(2.09 J/g°C)(0 - (-10.0°C))

Next, we calculate the heat lost by the water:

Qwater = mcΔT

where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The specific heat capacity of water is 4.18 J/g°C. Substitute the given values into the equation:

Qwater = (110.0 g)(4.18 J/g°C)(85.0°C - 0)

Now, we can solve for the final temperature by setting Qice + Qwater = 0:

(18.5 g)(2.09 J/g°C)(0 - (-10.0°C)) + (110.0 g)(4.18 J/g°C)(85.0°C - 0) = 0

Solving for the final temperature gives:

Final temperature = (Qice + Qwater) / (micecice + mwatercwater)

Final temperature = (-8.9525 kJ + 38957 J) / (18.5 g * 2.09 J/g°C + 110.0 g * 4.18 J/g°C)

Final temperature = 19.22 °C