In a physics experiment, 250g of water is placed in a 50g calorimeter cup made of aluminum when both are at 15C. Steam at 100C is completely absorbed by passing it through the water, raising the temperature of water and cup to 45C. Find the mass of steam absorbed. Specific heat of aluminum is 0.215 cal/gC

To find the mass of steam absorbed, we need to consider the transfer of heat between the steam and the water-calorimeter system. We can use the principle of conservation of energy, which states that the heat gained by an object must equal the heat lost by another object.

We can break down the problem into two separate heat transfer processes: one between the steam and water, and the other between the water and the aluminum cup.

Let's calculate the heat transfer for each process.

1. Heat transfer between steam and water:
The heat gained by the water can be calculated using the equation:
Q = m * c * ΔT
where Q is the heat gained, m is the mass of water, c is the specific heat capacity of water (1 cal/g°C), and ΔT is the change in temperature.

Plugging in the given values:
Q1 = (250g) * (1 cal/g°C) * (45°C - 15°C)
Q1 = 7500 cal

Since the steam transfers its heat to the water, the steam must lose the same amount of heat. Therefore, Q1 is also the heat lost by the steam.

2. Heat transfer between water and aluminum cup:
The heat gained by the water and aluminum cup can be calculated using the equation:
Q = m * c * ΔT
where Q is the heat gained, m is the combined mass of water and cup, c is the specific heat capacity of aluminum (0.215 cal/g°C), and ΔT is the change in temperature.

Plugging in the given values:
Q2 = (250g + 50g) * (0.215 cal/g°C) * (45°C - 15°C)
Q2 = 1075 cal

Since the water and cup gain heat, the heat must have been lost by the steam. Therefore, Q2 is also the heat lost by the steam.

Now, we can equate the two heat transfer equations to find the mass of steam absorbed:
Q1 = Q2

7500 cal = 1075 cal

To isolate the mass of steam (let's say msteam), we can rearrange the equation:

msteam * csteam * ΔT = (250g + 50g) * caluminum * ΔT

msteam = [(250g + 50g) * caluminum * ΔT] / csteam

Plugging in the known values:
msteam = [(250g + 50g) * (0.215 cal/g°C) * (45°C - 15°C)] / (1 cal/g°C)

msteam = 1075g

Therefore, the mass of steam absorbed is 1075 grams.