Question

In a reaction in a calorimeter the temperature of 32.0 grams of water increases from 19.50°C to 22.83°C. What is the amount of heat absorbed by the water?

Answer 1

445.8 J of heat is absorbed by the water

Step-by-step explanation:

Step 1: Data given

Mass of water = 32.0 grams

Initial temperature = 19.50 °C

Final temperature = 22.83 °C

Specific heat of water = 4.184 J.g°C

Step 2: Calculate the amount of heat absorbed by the water

Q = m*c*ΔT

⇒with Q = the heat absorbed = TO BE DETERMINED

⇒with m= the mass of water = 32.0 grams

⇒with c= the specific heat of water = 4.184 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = 22.83 °C - 19.50 °C = 3.33 °C

Q = 32.0 * 4.184 * 3.33

Q = 445.8 J

445.8 J of heat is absorbed by the water