Question

A vessel of negligible heat capacity is having volume $V_0$ of water at a temperature of $T_0$. From a faucet above water is falling from a geyser, at constant rate 'r' (volume flow rate). The temperature of water from geyser is $3T_0$. The CORRECT statements are

• A. The temperature of water in the vessel increases linearly with time
• B. The temperature of the water in the vessel increases but not linearly with-time
• C. The volume of water in vessel becomes $2V_0$ when the temperature $2T_0$
• D. The temperature of water is $\frac{5T_0}{3}$ after $t=\frac{V_0}{2r}$

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

Let $V(t)$ be the volume of water in the vessel at time $t$, and $T(t)$ be its temperature. Initially, $V(0) = V_0$ and $T(0) = T_0$. Water is added at a constant volume flow rate $r$ with a temperature of $3T_0$. The volume of water in the vessel at time $t$ is $V(t) = V_0 + rt$. The mass of water in the vessel is $M(t) = \rho V(t) = \rho (V_0 + rt)$, where $\rho$ is the density of water. The rate of mass entering the vessel is $\frac{dM}{dt} = \rho r$.

The rate of heat energy entering the vessel is given by the mass flow rate multiplied by the specific heat capacity $c$ and the temperature of the incoming water: Rate of heat input = $(\rho r) c (3T_0)$.

The total heat energy of the water in the vessel at time $t$ is $Q(t) = M(t) c T(t)$. The rate of change of heat energy in the vessel is $\frac{dQ}{dt} = \frac{d}{dt}(M(t) c T(t))$. Assuming no heat loss to the surroundings and negligible heat capacity of the vessel, the rate of heat input equals the rate of change of heat energy in the vessel: $\frac{d}{dt}(M(t) c T(t)) = (\rho r) c (3T_0)$ $c \frac{d}{dt}(\rho (V_0 + rt) T(t)) = 3 \rho r c T_0$ $\frac{d}{dt}((V_0 + rt) T(t)) = 3 r T_0$

Integrate both sides with respect to $t$: $(V_0 + rt) T(t) = \int 3 r T_0 dt = 3 r T_0 t + C$

Using the initial condition $T(0) = T_0$ at $t=0$: $(V_0 + r \cdot 0) T_0 = 3 r T_0 \cdot 0 + C$ $V_0 T_0 = C$

Substitute $C$ back into the equation: $(V_0 + rt) T(t) = 3 r T_0 t + V_0 T_0$ $T(t) = \frac{3 r T_0 t + V_0 T_0}{V_0 + rt}$ $T(t) = T_0 \frac{3 (rt/V_0) + 1}{1 + (rt/V_0)}$

Analysis of statements:

(A) The temperature of water in the vessel increases linearly with time The expression for $T(t)$ is $T(t) = T_0 \frac{1 + 3(rt/V_0)}{1 + (rt/V_0)}$. This is not a linear function of $t$. As $t \to \infty$, $T(t) \to 3T_0$, showing asymptotic behavior, not linear. The rate of temperature increase $\frac{dT}{dt} = \frac{2 T_0 r/V_0}{(1 + rt/V_0)^2}$ is not constant. Thus, statement (A) is incorrect.

(B) The temperature of the water in the vessel increases but not linearly with-time For $t > 0$, $rt/V_0 > 0$. $T(t) = T_0 \frac{1 + 3(rt/V_0)}{1 + (rt/V_0)}$. Since $1 + 3(rt/V_0) > 1 + (rt/V_0)$ for $rt/V_0 > 0$, $T(t) > T_0$. So the temperature increases. As shown in (A), the temperature increase is not linear. Thus, statement (B) is correct.

(C) The volume of water in vessel becomes $2V_0$ when the temperature $2T_0$ The volume becomes $2V_0$ when $V(t) = V_0 + rt = 2V_0$, which implies $rt = V_0$. Let's find the temperature at $rt = V_0$: $T(t) = T_0 \frac{1 + 3(rt/V_0)}{1 + (rt/V_0)} = T_0 \frac{1 + 3(V_0/V_0)}{1 + (V_0/V_0)} = T_0 \frac{1+3}{1+1} = T_0 \frac{4}{2} = 2T_0$. So, when the volume is $2V_0$, the temperature is $2T_0$. Statement (C) is correct.

(D) The temperature of water is $\frac{5T_0}{3}$ after $t=\frac{V_0}{2r}$ Let $t = \frac{V_0}{2r}$. Then $rt = r \cdot \frac{V_0}{2r} = \frac{V_0}{2}$. Substitute this into the temperature equation: $T(t) = T_0 \frac{1 + 3(rt/V_0)}{1 + (rt/V_0)} = T_0 \frac{1 + 3(V_0/2 / V_0)}{1 + (V_0/2 / V_0)} = T_0 \frac{1 + 3/2}{1 + 1/2} = T_0 \frac{5/2}{3/2} = \frac{5T_0}{3}$. So, at $t=\frac{V_0}{2r}$, the temperature is $\frac{5T_0}{3}$. Statement (D) is correct.

The correct statements are (B), (C), and (D).