If 60 kJ of heat energy are added to 2 kg of water at 30°C, what will be the final temperature of the water? The specific heat of water is 4186 J/kg·°C
60kJ = 60,000J
2kg = 2000g
Put this into q = m * C * delta t where delta t = (final temperature - initial temperature)
Let x = the final temperature of the water
60,000J = 2000g * (4.186J/(kg * ºC)) * (x - 30ºC)
Solve for x and you get x = 37.167ºC (I didn’t use proper significant figures here)
Hope that helped!