Answer to Question #66453 in General Chemistry for Kasha

Question

How much heat does it take to increase the temperature of 1.45 L of water from 0 deg Celsius to 100 deg Celsius? Assume water has a density of 1 g/mL

Answer 1

Solution:
Specific heat capacity is given by formula
S=(q÷(m×∆T))S=(q÷(m×∆T)),
Where:
s = specific heat capacity (sometimes represented by the letter c, or Cs)
q = heat
m = mass (we can calculate mass of water, using equation: m = ρ∙V = 1g/mL∙1450mL = 1450 (g))
Δ T = change in temperature: ∆T = 100 - 0 = 100.
As per Specific Heat Capacity of Water,
4.184 J of heat is required to raise temperature of 1 g of water by 1 ℃. So, the value of heat is:
q = S∙m∙∆T = 4.184∙1450∙100 = 606680 (J) = 606.68 (kJ)

Answer: q = 606.68 kJ.