Calculate the amount of heat required to heat a 3.3 kg gold bar from 26 ∘c to 69 ∘c. specific heat capacity of gold is 0.128 j/g∘c.

We need 18.2 kJ of heat

Explanation:

Step 1: Data given

Mass of gold = 3.3 kg = 3300 grams

Initial temperature = 26.0 °C

Final temperature = 69.0 °C

Specific heat of gold = 0.128 J/g°C

Step 2: Calculate the heat

Q = m*c*ΔT

⇒ with Q = the heat transfered = TO BE DETERMINED

⇒ with m = the mass of gold = 3300 grams

⇒ with c = the specific heat of gold = 0.128 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = 69.0 - 26.0 = 43.0 °C

Q = 3300g * 0.128 J/g°C * 43.0 °C

Q = 18163 J = 18.2 kJ

We need 18.2 kJ of heat

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o