what is the thermal energy change as 175.0g, of water drops from 75.00 dgrs C to 15.50 dgrs C???
q = mass H2O x specific heat water x (Tfinal - Tinitial).
mass H2O = 175.0 g
specific heat water = 4.184 J/g*C
Tf = 15.50
Ti = 75.00
Solve for q.
Check my work for typos.
Thermal energy= mass*specificheat*(Tfinal-Tinitial)
To find the thermal energy change, you can use the formula: Thermal energy = mass * specific heat * (Tfinal - Tinitial).
Given values:
Mass of water (m) = 175.0 g
Specific heat of water (C) = 4.184 J/g*C
Final temperature (Tfinal) = 15.50 degrees Celsius
Initial temperature (Tinitial) = 75.00 degrees Celsius
Now, substitute the given values into the formula:
Thermal energy = 175.0 g * 4.184 J/g*C * (15.50 degrees Celsius - 75.00 degrees Celsius)
Once you have substituted the values, you can simplify the expression:
Thermal energy = 175.0 g * 4.184 J/g*C * (-59.50 degrees Celsius)
Now, perform the calculation:
Thermal energy = -43927 J
Therefore, the thermal energy change as 175.0 g of water drops from 75.00 degrees Celsius to 15.50 degrees Celsius is -43927 J.