How much heat, in must be removed from 23.0 g of water at 67.2°C to decrease its temperature to 6.2°C . Please provide your final answer rounded to two decimal places

Answer 1

q = mC∆T

q = heat = ?

m = mass of water = 23.0 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 67.2º - 6.2º = 61º

Solving for q, we have...

q = (23.0 g)(4.184 J/gº)(61º)

q = 5870.15 J = 5.87 kJ (note: the sign would be negative since heat is being removed from the water)