How many joules of heat are given off when 5.00g of water cool from 348.0K to 298.0K?

Q =  -1045 J

Explanation:

Given data:

Mass of water = 5.00 g

Initial temperature = 348.0 K

Final temperature = 298.0 K

Heat given off = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g.K

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 298.0 K - 348.0 K

ΔT = - 50 K

Q = 5.0 g ×4.18 J/g.K× - 50 K

Q =  -1045 J

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.