what mass of steam intially at 130 degree celcius is needed to warm 200 grams of water in a 100 gram container from 20 to 50 degree celcius
To find the mass of steam required to warm the water, we need to use the equation:
Q = m * c * ΔT
Where:
Q is the heat energy transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's find the heat energy required to warm the water from 20 to 50 degrees Celsius. We'll assume the specific heat capacity of water is 4.18 J/g°C.
Q_water = m_water * c_water * ΔT_water
Q_water = 200 g * 4.18 J/g°C * (50°C - 20°C)
Q_water = 200 g * 4.18 J/g°C * 30°C
Q_water = 25080 J
Next, let's find the heat energy that the steam releases when it cools down from 130 to 100 degrees Celsius. The specific heat capacity of steam is about 2.03 J/g°C.
Q_steam = m_steam * c_steam * ΔT_steam
Q_steam = m_steam * 2.03 J/g°C * (130°C - 100°C)
Q_steam = m_steam * 2.03 J/g°C * 30°C
Q_steam = 60.9 m_steam J
The heat energy lost by the steam (Q_steam) must be equal to the heat energy absorbed by the water (Q_water).
So, 60.9 m_steam J = 25080 J
Now, isolate m_steam to solve for the mass of steam required:
m_steam = 25080 J / 60.9 J/g
m_steam ≈ 411.1 g (rounded to one decimal place)
Therefore, approximately 411.1 grams of steam initially at 130 degrees Celsius is needed to warm 200 grams of water in a 100 gram container from 20 to 50 degrees Celsius.