Answer:
The water will absorb 1004.16 Joule of heat
Explanation:
Step 1: Data given
Mass of the metal = 58.00 grams
Temperature of the metal = 100.00 °C
Mass of water = 60.00 grams
Temperature of water = 18.00 °C
Final temperature = 22.00 °C
Specific heat of water = 4.184 J/g°C
Step 2: Calculate the amount of heat absorbed by the water in joules
Q = mass *specific heat *ΔT
⇒ with Q = the heat absorbed by water
⇒ with mass of water = 60.00 grams
⇒ with specific heat of water = 4.184 J/g°C
⇒ with ΔT = The change in temperature of water = T2 - T1 = 22 - 18 = 4.0 °C
Q = 60.00 * 4.184 J/g°C * 4.0 °C
Q = 1004.16 J
The water will absorb 1004.16 Joule of heat
