Question

If the specific heat of a solution is 4.18 J/goC, and you have 296 mL (1.03 g/mL) which increases in temperature by 6.9 degrees, how many kilojoules did the solution absorb?

Answer 1

Answer:

Q = 8.8 kJ

Explanation:

Step 1: Data given

The specific heat of a solution = 4.18 J/g°C

Volume = 296 mL

Density = 1.03 g/mL

The temperature increases with 6.9 °C

Step 2: Calculate the mass of the solution

mass = density * volume

mass = 1.03 g/mL * 296 mL

mass = 304.88 grams

Step 3: Calculate the heat

Q = m*c*ΔT

⇒ with Q = the heat in Joules = TO BE DETERMINED

⇒ with m = the mass of the solution = 304.88 grams

⇒ with c = the specific heat of the solution = 4.18 J/g°C

⇒ with ΔT = the change in temperature = 6.9 °C

Q = 304.88 g * 4.18 J/g°c * 6.9 °C

Q = 8793.3 J = 8.8 kJ

Q = 8.8 kJ