Question

250g of glass at 60°C is placed in a container with 700g of alcohol. The specific heat capacities of glass and alcohol are 840J/kg°C and 2,500J/kg°C respectively. At thermal equilibrium, the final temperature of the mixture is 20°C. Determine the initial temperature of the alcohol assuming that no heat is lost while both substances are mixed

Answer 1

The initial temperature of the alcohol is approximately 10.4°C.

Step-by-step explanation:

To find the initial temperature of the alcohol, we can use the principle of conservation of energy. The heat lost by the glass will be equal to the heat gained by the alcohol.

The formula for calculating heat is:

Q = m * c * ΔT

where:

Q = heat (in joules)

m = mass (in kg)

c = specific heat capacity (in J/kg°C)

ΔT = change in temperature (in °C)

Let the initial temperature of the alcohol be T.

Calculate the heat lost by the glass:

Q_glass = m_glass * c_glass * (T_initial_glass - T_final)

where:

m_glass = mass of the glass (250g = 0.25 kg)

c_glass = specific heat capacity of glass (840 J/kg°C)

T_initial_glass = initial temperature of the glass (60°C)

T_final = final temperature of the mixture (20°C)

Q_glass = 0.25 * 840 * (60 - 20)

Calculate the heat gained by the alcohol:

Q_alcohol = m_alcohol * c_alcohol * (T_final - T_initial_alcohol)

where:

m_alcohol = mass of the alcohol (700g = 0.7 kg)

c_alcohol = specific heat capacity of alcohol (2,500 J/kg°C)

T_initial_alcohol = initial temperature of the alcohol (unknown)

Q_alcohol = 0.7 * 2500 * (20 - T_initial_alcohol)

Since no heat is lost while both substances are mixed, Q_glass = Q_alcohol:

0.25 * 840 * (60 - 20) = 0.7 * 2500 * (20 - T_initial_alcohol)

Now, solve for T_initial_alcohol:

16800 = 1750 * (20 - T_initial_alcohol)

Divide both sides by 1750:

(20 - T_initial_alcohol) = 16800 / 1750

(20 - T_initial_alcohol) = 9.6

Subtract 9.6 from both sides:

T_initial_alcohol = 20 - 9.6

T_initial_alcohol = 10.4°C

Therefore, the initial temperature of the alcohol is approximately 10.4°C.