What is the mass of water that can be heated from 39 C to 80 C by the combustion of 150 g of methane. CH4 + 2O2 => CO2 + 2H2O Delta H= -890kJ
To calculate the mass of water that can be heated from 39°C to 80°C by the combustion of 150 g of methane, we need to use the concept of heat transfer.
First, let's calculate the heat released by the combustion of methane. The given equation for the combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O (ΔH = -890 kJ)
The negative sign in front of the 890 kJ indicates that the reaction is exothermic, meaning it releases heat.
Since we know the combustion of 150 g of methane is exothermic and releases 890 kJ of heat, we can calculate the amount of heat released per gram of methane:
Heat released per gram of methane = 890 kJ / 150 g
Next, we need to calculate the heat required to increase the temperature of water. The specific heat capacity of water is approximately 4.18 J/g°C. This means it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
Now, we need to calculate the heat required to increase the temperature of water from 39°C to 80°C. The temperature change is 80°C - 39°C = 41°C.
Heat required to increase the temperature of water = specific heat capacity of water × mass of water × temperature change
Since we know that 1 kJ = 1000 J, we need to convert the heat required to increase the temperature of water from kJ to J.
Finally, we can calculate the mass of water by rearranging the equation and solving for mass:
Mass of water = (Heat released by combustion of methane) / (heat required to increase the temperature of water)
Using these calculations, we can determine the mass of water that can be heated from 39°C to 80°C by the combustion of 150 g of methane.