Solution
1. Mathematical Formulation
Let the temperature of the hot water reservoir be $T_R$.
Let the initial mass of water in the calorimeter be $m_0 = 1.0 \text{ kg}$ at initial temperature $T_0 = 20^\circ\text{C}$.
Hot water is added at a constant rate $\mu = \frac{dm}{dt}$ (kg/s).
At any time $t$, the total mass of water in the calorimeter is $m(t) = m_0 + \mu t$.
Using the principle of conservation of energy (assuming the specific heat capacity of water $c$ is constant), the total enthalpy of the mixed water must equal the sum of the enthalpy of the initial water and the enthalpy of the added hot water:
$$ (m_0 + \mu t) c \theta = m_0 c T_0 + (\mu t) c T_R $$Canceling the specific heat $c$ and rearranging to solve for the mixture temperature $\theta$:
$$ \theta = \frac{m_0 T_0 + \mu t T_R}{m_0 + \mu t} $$Rearranging to isolate unknowns: $$ (m_0 + \mu t)\theta = m_0 T_0 + \mu t T_R $$ $$ m_0(\theta – T_0) = \mu t (T_R – \theta) $$
2. Analyzing Graph Data Points
From the provided graph, we can identify two distinct coordinates $(t, \theta)$:
- Point A: $t_1 = 200 \text{ s}$, $\theta_1 = 30^\circ\text{C}$
- Point B: $t_2 = 500 \text{ s}$, $\theta_2 = 40^\circ\text{C}$
3. Solving for Reservoir Temperature ($T_R$)
Substituting Point A into our energy balance equation:
$$ m_0(30 – 20) = \mu (200) (T_R – 30) $$ $$ 10 m_0 = 200 \mu (T_R – 30) \implies \frac{m_0}{\mu} = 20(T_R – 30) \quad \text{— (i)} $$Substituting Point B into the equation:
$$ m_0(40 – 20) = \mu (500) (T_R – 40) $$ $$ 20 m_0 = 500 \mu (T_R – 40) \implies \frac{m_0}{\mu} = 25(T_R – 40) \quad \text{— (ii)} $$Equating the values of $\frac{m_0}{\mu}$ from (i) and (ii):
$$ 20(T_R – 30) = 25(T_R – 40) $$ $$ 4(T_R – 30) = 5(T_R – 40) $$ $$ 4T_R – 120 = 5T_R – 200 $$ $$ T_R = 200 – 120 $$ $$ T_R = 80 $$