Solution: Flow Calorimetry with Melting Ice
1. Principle of Energy Conservation
We analyze the steady-state flow of water through the container.
Incoming: Water at rate $r_i = 3.0 \text{ g/s}$ at temperature $\theta_i = 28^{\circ}\text{C}$.
Inside: The warm water melts some ice. The ice is at $\theta_{ice} = 0^{\circ}\text{C}$.
Outgoing: The mixture (incoming water + melted ice) leaves at $\theta_0 = 1.0^{\circ}\text{C}$.
2. Heat Balance Equation
Heat lost by incoming water = Heat gained by melting ice + Heat to warm melted ice to $\theta_0$.
Let the rate of melting ice be $r_{melt}$.
$$ \text{Heat Lost by Inlet} = r_i s (\theta_i – \theta_0) $$ $$ \text{Heat Gained by Ice} = r_{melt} L + r_{melt} s (\theta_0 – 0) $$Equating them:
$$ r_i s (\theta_i – \theta_0) = r_{melt} [ L + s \theta_0 ] $$ $$ r_{melt} = r_i \frac{s(\theta_i – \theta_0)}{L + s \theta_0} $$3. Calculation of Outlet Rate
The rate of water coming out, $r_{out}$, is the sum of the inlet rate and the melting rate.
$$ r_{out} = r_i + r_{melt} $$ $$ r_{out} = r_i \left( 1 + \frac{s(\theta_i – \theta_0)}{L + s \theta_0} \right) $$simplifying the fraction:
$$ r_{out} = r_i \left( \frac{L + s\theta_0 + s\theta_i – s\theta_0}{L + s\theta_0} \right) = r_i \left( \frac{L + s\theta_i}{L + s\theta_0} \right) $$Using the notation from the problem (where $\theta_{ice}=0$ is implicit):
$$ r_{out} = r_i \frac{L + s(\theta_i – 0)}{L + s(\theta_0 – 0)} $$Substitute values ($s = 4.2 \text{ J/g}^{\circ}\text{C}$, $L = 336 \text{ J/g}$):
$$ r_{out} = 3.0 \times \frac{336 + 4.2(28)}{336 + 4.2(1)} $$ $$ r_{out} = 3.0 \times \frac{336 + 117.6}{336 + 4.2} $$ $$ r_{out} = 3.0 \times \frac{453.6}{340.2} $$ $$ r_{out} = 3.0 \times 1.333… $$ $$ r_{out} = 4.0 \text{ g/s} $$