The heat of fusion of water is 335 J/g, the heat of vaporization of water is 2.26 kJ/g, and the specific heat of water is 4.184 J/deg/g. How many grams of ice at 0 ° could be converted to steam at 100 °C by 9,946 Joules of heat?
Let Y = grams ice.
335*Y + (Y*4.184*100) + 2226*Y = 9946.
Solve for Y in grams.
To determine how many grams of ice can be converted to steam by 9,946 Joules of heat, we need to consider the different stages of the process: heating the ice to 0 °C, melting the ice into water at 0 °C, heating the water from 0 °C to 100 °C, and finally, vaporizing the water into steam at 100 °C.
1. Heating the ice to 0 °C:
To calculate the heat required to raise the temperature of the ice, we need to know the mass of the ice and its specific heat capacity. In this case, we don't have the mass of the ice, but we can assume it as 'm' grams.
The heat required to raise the temperature of the ice to 0 °C is given by:
Heat1 = m * specific heat of ice * change in temperature
= m * 4.184 J/deg/g * (0 °C - (-273.15 °C))
Note: We subtract -273.15 °C to convert 0 °C to the Kelvin scale, which is used for specific heat of ice.
2. Melting the ice into water at 0 °C:
The heat required to melt the ice is given by:
Heat2 = m * heat of fusion of water
3. Heating the water from 0 °C to 100 °C:
The heat required to raise the temperature of the water from 0 °C to 100 °C is given by:
Heat3 = m * specific heat of water * change in temperature
= m * 4.184 J/deg/g * (100 °C - 0 °C)
4. Vaporizing the water into steam at 100 °C:
The heat required to vaporize the water is given by:
Heat4 = m * heat of vaporization of water
Now, to find the mass 'm' of the ice, we need to solve the equation:
9,946 J = Heat1 + Heat2 + Heat3 + Heat4
Let's plug in the known values to calculate the mass of the ice.