What quantity of heat is required to warm 53.7 g of liquid water from 0°C to 100°C?
q = mass x specific heat water x 100
To find the quantity of heat required to warm a substance, we can use the formula:
Q = m * c * ΔT
Where:
Q is the quantity of heat
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we are looking to find the quantity of heat required to warm 53.7 g of liquid water from 0°C to 100°C.
First, let's find the specific heat capacity of water. The specific heat capacity of water is 4.184 J/g°C, meaning it takes 4.184 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
Now, let's calculate the quantity of heat using the given information:
m = 53.7 g (mass of water)
c = 4.184 J/g°C (specific heat capacity of water)
ΔT = 100°C - 0°C = 100°C (change in temperature)
Plugging these values into the formula:
Q = 53.7 g * 4.184 J/g°C * 100°C
Calculating this:
Q = 22434.048 J
So, the quantity of heat required to warm 53.7 g of liquid water from 0°C to 100°C is approximately 22434.048 Joules.