Question

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.

Answer 1

We need 18.2 kJ of heat

Step-by-step 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

Answer 2

To answer the question above, we make use of the equation,
heat = (specific heat ) x mass x (change in temperature)
Substituting the known values,
heat = (0.128 J/g°C) x (3.3 kg)(1000g / 1kg) x (69°C - 26°C)
= 18163.2 J
Therefore, the amount of heat required is approximately 18163.2 J.