Question

A 1.0 kg sample of water at \( 80^\circ C \) is placed in thermal contact with a 2.0 kg sample of water at \( 20^\circ C \). If the system is insulated, what will be the final equilibrium temperature of the system? Assume no heat loss to the surroundings, and the specific heat capacity of water is \( 4.18 \, \text{J/g}^\circ \text{C} \).

• A. \( 40^\circ C \)
• B. \( 45^\circ C \)
• C. \( 50^\circ C \)
• D. \( 60^\circ C \)

Hint:

When two bodies of different temperatures come into thermal contact, the final temperature can be determined using the principle of conservation of energy. Remember that the heat lost by the hot body is equal to the heat gained by the cold body.

Answer 1

The Correct Option is B

To find the final temperature, we use the principle of conservation of energy, which states that the heat lost by the hotter sample is equal to the heat gained by the colder sample: \[ m_1 c \Delta T_1 = m_2 c \Delta T_2 \] Where: - \( m_1 = 1.0 \, \text{kg} = 1000 \, \text{g} \) (mass of the hot sample), - \( m_2 = 2.0 \, \text{kg} = 2000 \, \text{g} \) (mass of the cold sample), - \( c = 4.18 \, \text{J/g}^\circ \text{C} \) (specific heat capacity of water), - \( \Delta T_1 = T_{\text{final}} - 80 \) (change in temperature of the hot sample), - \( \Delta T_2 = T_{\text{final}} - 20 \) (change in temperature of the cold sample). Setting up the energy balance: \[ 1000 \times 4.18 \times (T_{\text{final}} - 80) = 2000 \times 4.18 \times (T_{\text{final}} - 20) \] Canceling the \( 4.18 \) term from both sides: \[ 1000 \times (T_{\text{final}} - 80) = 2000 \times (T_{\text{final}} - 20) \] Expanding both sides: \[ 1000 T_{\text{final}} - 80000 = 2000 T_{\text{final}} - 40000 \] Solving for \( T_{\text{final}} \): \[ 1000 T_{\text{final}} - 2000 T_{\text{final}} = -40000 + 80000 \] \[ -1000 T_{\text{final}} = 40000 \] \[ T_{\text{final}} = 40^\circ C \] Thus, the final equilibrium temperature is \( 45^\circ C \).