How much heat (kJ) is absorbed by 229.1 g of water in order for the temperature to increase from 25.00∘C to 32.50∘C?

Q2) Calculate the amount of heat required to raise the temperature of a 34 g sample of water from 9 ∘C to 23 ∘C.

(Q1) 9.42 kJ.

(Q2) 1.999 kJ

Explanation:

Heat: This is a form of Energy that brings about the sensation of warmth.

The S. I unit of Heat is Joules (J).

The heat of a body depend on the mass of the body, specific heat capacity, and temperature difference. as shown below

Q = cm (t₂-t₁) ... Equation 1

(Q1)

Q = cm (t₂-t₁)

Where Q = amount of heat absorbed, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 229.1 g = 0.2991 kg, t₁ = 25.0 °C, 32.50 °C

Constant: c = 4200 J/kg.°C

Substituting into equation 1

Q = 0.2991*4200 (32.5-25)

Q = 1256.22 (7.5)

Q = 9421.65 J

Q = 9.42 kJ.

Hence the heat absorbed = 9.42 kJ

(Q2)

Q = cm (t₂-t₁)

Where Q = amount of heat required, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 34 g = 0.034 kg, t₁ = 9 °C, t₂ = 23 °C

Constant: c = 4200 J/kg.°C

Q = 0.034*4200 (23-9)

Q = 142.8 (14)

Q = 1999.2 J

Q = 1.999 kJ.

Thus the Heat required = 1.999 kJ