A 350g glass beaker contains 500g of water at a temperature of 10.0°C. If 400g of ethyl alcohol at 35°C is subsequently poured into

the beaker and the water and alcohol mixture is thoroughly stirred, what is its equilibrium temperature?

The glass and the water start out at 10 C and get warmer as they gain heat from the ethyl alcohol, initially at 35 C. Look up the specific heats of alcohol, glass and water.

Write an equation that says the heat gained by the water and glass equals the heat lost by the alcohol. The only unknown should be the final temperature, which you can then solve for.

To find the equilibrium temperature of the water and ethyl alcohol mixture, we can use the principle of energy conservation. The total heat gained by the system must be equal to the total heat lost by the system.

First, let's calculate the heat gained by the water:

Q_water = m_water * c_water * ΔT_water

where:
- Q_water is the heat gained by the water
- m_water is the mass of water
- c_water is the specific heat capacity of water
- ΔT_water is the change in temperature of the water

The mass of water is 500g, the specific heat capacity of water is 4.18 J/g°C, and the initial temperature is 10.0°C. Since we want to find the equilibrium temperature, we need to subtract the initial temperature from the final equilibrium temperature to get the change in temperature:

ΔT_water = T_eq - T_initial = T_eq - 10.0°C

Next, let's calculate the heat gained by the ethyl alcohol:

Q_alcohol = m_alcohol * c_alcohol * ΔT_alcohol

The mass of ethyl alcohol is 400g, the specific heat capacity of ethyl alcohol is 2.44 J/g°C, and the initial temperature is 35°C. Again, we subtract the initial temperature from the final equilibrium temperature to get the change in temperature:

ΔT_alcohol = T_eq - T_initial = T_eq - 35.0°C

Now, since energy is conserved, the total heat gained by the water is equal to the total heat gained by the ethyl alcohol:

Q_water = Q_alcohol

Using the equations above, we can set up the following equation:

m_water * c_water * (T_eq - 10.0°C) = m_alcohol * c_alcohol * (T_eq - 35.0°C)

Substituting the known values:

500g * 4.18 J/g°C * (T_eq - 10.0°C) = 400g * 2.44 J/g°C * (T_eq - 35.0°C)

Simplifying and rearranging the equation:

2090(T_eq - 10.0°C) = 976(T_eq - 35.0°C)

2090T_eq - 20900.0J = 976T_eq - 34240.0J

2090T_eq - 976T_eq = -34240.0J + 20900.0J

1114T_eq = -13340.0J

T_eq = -13340.0J / 1114

T_eq ≈ -11.98°C

Therefore, the equilibrium temperature of the water and ethyl alcohol mixture is approximately -11.98°C.