What is the mass in grams of a sample of water when the temperature drops from 75.00∘C to 56.00∘C and 800.0 J of energy are lost?

Let y g be the mass of the sample of water.

Energy change = m c ΔT
800 J = (y g) × [4.184 J/(g°C)] × [(75.00 - 56.00)°C]
y = 800 / [4.184 × (75.00 - 56.00)]
y = 10.06

The mass of the sample of water = 10.06 g

The specific heat of water is: Water (liquid): CP = 4.186 J/(g⋅°C )
This means that when 4.186J of heat energy is added to 1g of water , the temperature will increase by 1°C
Conversely , when 1g of water loses 4.186J of heat energy , the temperature will drop by 1°C
We use this information to solve your problem :
Heat energy = Mass X specific heat X change in temperature
You know:
Heat energy: 800.0J
Mass of water = ??
Specific heat = 4.186J/(g°C)
Temerature change = 75.00 - 56.00 = 19°C
Now substitute and solve for mass
800.0J = Mass X 4.186 X 19
Mass = 800.0 /( 4.186 X 19 )
Mass = 10.0586g
You have used 4 significant digits in your question:
Answer: Mass = 10.06g