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

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation: