1)How much heat is required for 20.0 g of water to go from 23C to 38C?
2)Find the amount of energy needed to melt 15.0 g of Ice at 0 C
1. q = heat = mass H2O x specific heat H2O x (Tfinal-Tinitial)
2. q = mass ice x heat fusion
To answer both questions, we need to use the specific heat capacity and the enthalpy of fusion of water.
1) To calculate the heat required for water to go from one temperature to another, we use the formula:
Q = m * c * ΔT
Where:
Q = heat energy (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance
ΔT = change in temperature (in Celsius)
The specific heat capacity of water is approximately 4.184 J/g°C (This value represents the amount of heat energy required to raise the temperature of 1 gram of water by 1 degree Celsius.)
Using the formula, we can calculate the heat required for water:
Q = (20.0 g) * (4.184 J/g°C) * (38°C - 23°C)
Q = 20.0 g * 4.184 J/g°C * 15°C
Q = 1,252.8 J
Therefore, 1,252.8 Joules of heat is required for 20.0 g of water to go from 23°C to 38°C.
2) To calculate the energy needed to melt ice, we use the formula:
Q = m * ΔHf
Where:
Q = heat energy (in Joules)
m = mass of the substance (in grams)
ΔHf = enthalpy of fusion (heat of fusion) of the substance
The enthalpy of fusion for water is approximately 334 J/g (This value represents the amount of heat energy required to change 1 gram of ice at 0°C into 1 gram of liquid water at 0°C.)
Using the formula, we can calculate the energy needed to melt 15.0 g of ice:
Q = (15.0 g) * (334 J/g)
Q = 5,010 J
Therefore, 5,010 Joules of energy is needed to melt 15.0 g of ice at 0°C.