What is the change in heat energy when 40.0g of water at 25.0oc is raised to 60.0oc?

Given:

40.0g of water

25.0oC is raised to 60.0oC

Required:

Change in heat energy

Solution:

This can be solved through the equation H = mCpT where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.

The specific heat of the water is 4.18 J/g-K

Plugging in the values into the equation

H = mCpT

H = (40g) (4.18 J/g-K) (60 – 25)

H = 5,852 J

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i