In which case is heat involved: mixing 100.0-g samples of 90'C water and 80'C water or mixing 100.0-g samples of 60'C water? assume no heat is lost to the environment.

please help thanks.

In both cases, heat is involved.

When you mix the 100.0-g samples of 90°C water and 80°C water, heat will be transferred from the hotter water (90°C) to the colder water (80°C) until they reach a thermal equilibrium. Therefore, heat will flow from the 90 °C water to the 80 °C water.

Similarly, when you mix the 100.0-g samples of 60°C water, heat will also be involved. In this case, heat will flow from the 60°C water to the surrounding environment until they reach a thermal equilibrium. However, since you assumed no heat is lost to the environment, the temperature of the mixture will reach a value between 60°C and the initial temperature of the surroundings - most likely somewhere around 60°C.

So, in both cases, heat transfer occurs during the mixing process.

To determine the case in which heat is involved, we need to compare the final temperature of the mixture with the initial temperatures of the two samples being mixed.

In the first case, where we mix 100.0 g of 90°C water and 100.0 g of 80°C water, we can assume that heat is exchanged between the two samples until they reach a common final temperature. Since both water samples are initially above room temperature (assumed to be 25°C), the heat exchange will occur until the final temperature is reached.

In the second case, where we mix 100.0 g of 60°C water, we need to consider whether this water sample is above or below the room temperature. If it is above room temperature, heat will be lost to the environment until the final temperature is reached. If it is below room temperature, heat will be gained from the environment until the final temperature is reached.

Now let's calculate the final temperature for each case using the principle of energy conservation:

1. Case 1: Mixing 100.0 g of 90°C water and 100.0 g of 80°C water
Total heat lost by 90°C water = mcΔT = (100.0 g)(4.18 J/g°C)(90°C - Tfinal)
Total heat gained by 80°C water = mcΔT = (100.0 g)(4.18 J/g°C)(Tfinal - 80°C)
Here, 4.18 J/g°C is the specific heat capacity of water.

By assuming no heat loss to the environment, we can equate the heat lost by the hot water to the heat gained by the cold water:

(100.0 g)(4.18 J/g°C)(90°C - Tfinal) = (100.0 g)(4.18 J/g°C)(Tfinal - 80°C)

Simplifying the equation:
3762(Tfinal) - 3762(Tfinal)^2 - 33480 - 33480 = 33480(Tfinal) - 33480(80)
3762(Tfinal)^2 + 30072(Tfinal) - 66960 = 0

Solving this quadratic equation gives us two possible values for Tfinal: 10.2°C and 80.0°C.

This means if the final temperature is 10.2°C, then heat will be lost to the environment, but if the final temperature is 80.0°C, then no heat will be lost to the environment.

2. Case 2: Mixing 100.0 g of 60°C water
Total heat lost by 60°C water = mcΔT = (100.0 g)(4.18 J/g°C)(60°C - Tfinal)
Here, again 4.18 J/g°C is the specific heat capacity of water.

Similarly, by assuming no heat loss to the environment, we can equate the heat lost by the hot water to the heat gained from the environment:

(100.0 g)(4.18 J/g°C)(60°C - Tfinal) = (100.0 g)(4.18 J/g°C)(Tfinal - 25°C)

Simplifying the equation:
4180(Tfinal) - 4177(Tfinal)^2 - 6270 = 4178(Tfinal) - 104450
4177(Tfinal)^2 + 2(Tfinal) - 98360 = 0

Solving this quadratic equation gives us two possible values for Tfinal: 11.5°C and -8.8°C.

This means if the final temperature is 11.5°C, then heat will be gained from the environment, but if the final temperature is -8.8°C, then heat will be lost to the environment.

In conclusion, heat is involved in both cases: mixing 100.0 g samples of 90°C water with 80°C water, and mixing 100.0 g samples of 60°C water.

Mixing 100 g sample of the same T has no heat exchange.

Mixing two different temperataures has a heat exchange. One gets warmer; the other gets cooler.