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A coffee maker makes coffee by passing steam through a mixture of coffee powder, milk and water. If the steam is mixed at the rate of $50 \,g$ per minute in a mug containing $500 \,g$ of mixture, then it takes about $t_0$ seconds to make coffee at $70^{\circ} C$ when the initial temperature of the mixture is $25^{\circ} C$. The value of $t_0$ is close to .......... $s$ (ratio of latent heat of evaporation to specific heat of water is $540^{\circ} C$ and specific heat of the mixture can be taken to be the same as that of water)
$30$
$45$
$60$
$90$
Solution
(b)
Let $m$ gram of steam is condensed in the process of heating mixture from $25^{\circ} C$ to $70^{\circ} C$.
Then,
Heat lost by steam = Heat gained by mixture
$\Rightarrow$ Heat of condensation of steam + Heat given by water formed = Heat gained by mixture
$\Rightarrow m \cdot L+m s_w \Delta T=M \cdot s_m \Delta T$
$\Rightarrow m L+m s_w(100-70)=500 \times s_w \times(70-25)$
$\Rightarrow m=\frac{500 \times s_w \times 45}{L+30 s_w}$
$\Rightarrow m=\frac{500 \times 45}{\left(\frac{L}{s_w}+30\right)}$
$\Rightarrow m=\frac{500 \times 45}{(540+30)} \approx 40 g$
Now, in $1 \,min , 50 \,g$ of steam is condensed.
$\therefore 40 g$ of steam will be condensed in time $t_0$,
$t_0=\frac{40 \times 60}{50}- s =48 \,s$
Nearest answer is $45 \,s$.
