How much heat is released by the burning of a nut that causes a cup containing 50g of water to experience an increase in temperature from 25 degrees celsius to 35 degrees celsius

There is 2092 J of heat released

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

2093 J is released by the burning of the nut

Explanation:

Calorimetry formula: Q = m. C.ΔT

Q = 50g. C.ΔT

C is the specific heat of water: 4.186 J/g°C

ΔT = Final temperature - Initial temperature

ΔT = 35° - 25° = 10°C

Q = 50g. 4.186 J/g°C. 10°C → 2093 joules