A mass of 2 kg of water at a temperature of $18^{\circ}C$ is poured into a well-insulated container at a temperature of $15^{\circ}C$. The temperatures of the water and the container reach equilibrium at $17.4^{\circ}C$. Determine the amount of heat transferred and its conventional direction, when the system considered is: (a) the container with the insulation, (b) the water, and (c) the container with the insulation and the water. The specific heat of water is 1 kcal/kgK.

1. Problem Description

A mass of 2 kg of water at a temperature of $18^{\circ}C$ is poured into a well-insulated container at a temperature of $15^{\circ}C$. The temperatures of the water and the container reach equilibrium at $17.4^{\circ}C$. Determine the amount of heat transferred and its conventional direction, when the system considered is: (a) the container with the insulation, (b) the water, and (c) the container with the insulation and the water. The specific heat of water is 1 kcal/kgK.

2. Solution Steps

a) For the container:

Let $m_w$ be the mass of water, $T_{wi}$ be the initial temperature of water, $T_{ci}$ be the initial temperature of the container, $T_f$ be the final temperature of the mixture, and $c_w$ be the specific heat of water.

$m_w = 2 \, kg$

$T_{wi} = 18^{\circ}C$

$T_{ci} = 15^{\circ}C$

$T_f = 17.4^{\circ}C$

$c_w = 1 \, kcal/kgK$

Let $Q_c$ be the heat gained by the container.

The container gains heat from the water. Therefore, the heat gained by the container is positive.

$Q_c = -Q_w$

$Q_w = m_w c_w (T_f - T_{wi})$

$Q_w = 2 \, kg \times 1 \, kcal/kgK \times (17.4^{\circ}C - 18^{\circ}C)$

$Q_w = 2 \, kg \times 1 \, kcal/kgK \times (-0.6 \, K)$

$Q_w = -1.2 \, kcal$

$Q_c = -Q_w = -(-1.2) \, kcal = 1.2 \, kcal$

The container gains $1.2 \, kcal$.

b) For the water:

The water loses heat to the container. Therefore, the heat lost by the water is negative.

$Q_w = m_w c_w (T_f - T_{wi})$

$Q_w = 2 \, kg \times 1 \, kcal/kgK \times (17.4^{\circ}C - 18^{\circ}C)$

$Q_w = 2 \, kg \times 1 \, kcal/kgK \times (-0.6 \, K)$

$Q_w = -1.2 \, kcal$

The water loses $1.2 \, kcal$.

c) For the container and the water:

Since the system is well-insulated, there is no heat exchange with the surroundings. Therefore, the total heat change is zero.

$Q_{total} = Q_w + Q_c = 0$

$Q_{total} = -1.2 \, kcal + 1.2 \, kcal = 0$

3. Final Answer

a) The container gains $1.2 \, kcal$.

b) The water loses $1.2 \, kcal$.

c) The container with the water has no heat exchange with the surrounding. The total heat transfer is 0 kcal.