You add 7800 calories of heat to a block of ice at -40 degrees celsius. the temperature of the ice rises to 0 degrees celsius. how much ice did you heat up?
q = mass ice x specific heat ice x delta T.
q = 7800 calories.
mass = unknown
you will need to look up specific heat solid ice.
delta T = 40.
To determine how much ice you heated up, you need to calculate the amount of heat energy required to raise the temperature of the ice from -40 degrees Celsius to 0 degrees Celsius.
The specific heat capacity of ice is 2.09 J/g°C, which means it requires 2.09 joules of heat energy to raise the temperature of 1 gram of ice by 1 degree Celsius.
First, we need to find the mass of the ice. Let's assume the mass of the ice is "m" grams.
The formula to calculate the heat energy is:
Q = m * c * ΔT
Where:
Q is the heat energy in joules,
m is the mass of the substance in grams,
c is the specific heat capacity of the substance in J/g°C, and
ΔT is the change in temperature.
Given:
Q = 7800 calories
c = 2.09 J/g°C
ΔT = (0°C - (-40°C)) = 40°C
We need to convert the calories to joules:
1 calorie = 4.184 joules
7800 calories * 4.184 joules/calorie = 32652.8 joules
Now we can rearrange the formula to solve for mass:
m = Q / (c * ΔT)
m = 32652.8 joules / (2.09 J/g°C * 40°C)
m ≈ 391.3 grams
Therefore, you heated up approximately 391.3 grams of ice.