Liquid oxygen boils at a chilly -183.0o C when the pressure is one atmosphere. A small silver disk of mass 34.0 grams and temperature 27.3o C is dropped into the liquid. What mass of oxygen boils off as the coin cools to -183.0o C?

I need ther answer and the eequation thanks!

To solve this problem, we need to use the heat transfer equation. The change in heat (Q) can be calculated using the formula:

Q = m * c * ΔT

Where:
- Q is the heat transferred,
- m is the mass of the substance,
- c is the specific heat capacity, and
- ΔT is the change in temperature.

In this case, we need to find the mass of the oxygen that boils off, so we can rearrange the formula as follows:

m = Q / (c * ΔT)

Given:
- m (mass of the silver disk) = 34.0 grams
- ΔT (change in temperature) = -183.0°C - 27.3°C = -210.3°C (we convert to Celsius)

Now we need to find the Q value, which represents the heat transferred. To do this, we consider that the heat transferred from the silver disk is equal to the heat absorbed by the liquid oxygen. The heat transferred can be calculated using the formula:

Q = m * c * ΔT

Given:
- m (mass of the silver disk) = 34.0 grams
- c (specific heat capacity of silver) = 0.235 J/g°C
- ΔT (change in temperature) = -210.3°C

Now we can substitute these values into the equation:

Q = (34.0 g) * (0.235 J/g°C) * (-210.3°C)

Calculating this, we find:

Q ≈ -1642.17 J

Finally, we can substitute the calculated Q value and the given values for the liquid oxygen into the equation to determine the mass that boils off:

m = Q / (c * ΔT)

m = -1642.17 J / (0.918 J/g°C * -210.3°C)

Simplifying this, we find the mass of oxygen boiled off:

m ≈ -1642.17 J / (-196.68 J/g)

m ≈ 8.35 grams

Therefore, approximately 8.35 grams of oxygen will boil off as the silver disk cools to -183.0°C.