100 ml of water is initially at 20°C. 30,000 J of heat is added to the water. What is temperature change for the water?

Therefore the new temperature of the water is 91.67°C

Explanation:

Specific heat of water = 4.186 joule / gram °C

The equation to calculate change of temperature

Q=s*m*ΔT

Q = the amount of heat = 30,000 J

s = Specific heat of water = 4.186 joule / gram °C

m = mass of the water = volume * density = (100*1) gram = 100 gram

ΔT = the change of temperature = (T-20)

Therefore,

30000 = 4.186*100 * (T-20)

⇔ (T-20) = 71.67

⇔T = 71.67+20

⇔T = 91.67

Therefore the new temperature of the water is 91.67°C