Answer:
There is 2092 J of heat released
Step-by-step explanation:
Step 1: Data given
Mass of the water = 50.0 grams
Initial temperature = 25.0 °C
Final temperature = 35.0 °C
Specific heat of water = 4.184 J/g°C
Step 2: Calculate the heat released
Q = m*c*ΔT
⇒ m = the mass of water = 50.0 grams
⇒ c = the specific heat of water = 4.184 J/g°C
⇒ ΔT = The change of temperature = T2 - T1 = 35.0 - 25.0 = 10.0 °C
Q = 50.0g * 4.184 J/g°C * 10.0°C
Q = 2092 J
There is 2092 J of heat released