In Newton's experiment of cooling,the water e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T - T_0)$.For a calorimeter,the rate of heat loss is $\fra

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(C) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T - T_0)$.For a calorimeter,the rate of heat loss is $\frac{dQ}{dt} = (mC + W) \frac{dT}{dt}$,where $m$ is the mass,$C$ is the specific heat,and $W$ is the water equivalent.Since the cooling range $(70^{\circ}C$ to $60^{\circ}C)$ and the surrounding temperature are the same,the rate of heat loss $\frac{dQ}{dt}$ is constant for both.Thus,$\frac{(m_W C_W + W) \Delta T}{t_W} = \frac{(m_L C_L + W) \Delta T}{t_L}$.Rearranging for $C_L$: $C_L = \frac{1}{m_L} \left[ \frac{t_L}{t_W} (m_W C_W + W) - W ight]$.Given: $m_W = 350 \ g$,$C_W = 1 \ Cal/g^{\circ}C$,$W = 10 \ g$,$t_W = 3 \ min = 180 \ s$,$m_L = 300 \ g$,$t_L = 95 \ s$.Substituting the values: $C_L = \frac{1}{300} \left[ \frac{95}{180} (350 	imes 1 + 10) - 10 ight]$.$C_L = \frac{1}{300} \left[ \frac{95}{180} (360) - 10 ight] = \frac{1}{300} [95 	imes 2 - 10] = \frac{190 - 10}{300} = \frac{180}{300} = 0.6 \ Cal/g^{\circ}C$.

Similar Questions

$A$ container is filled with a liquid that cools from $100^{\circ}C$ to $70^{\circ}C$ in $5 \text{ min}$, when kept at a room temperature of $30^{\circ}C$. The time that it must have taken to cool down to $80^{\circ}C$ from its initial temperature is approximately: (in $\text{ min}$)

Two metallic spheres $P$ and $Q$ are made of the same material and have the same smoothness,but the weight of $P$ is $8$ times that of $Q$. If the two are heated to the same temperature and left to cool,then the ratio of the rate of cooling of $Q$ to that of $P$ is:

$A$ hollow copper sphere $S$ and a hollow copper cube $C$,both with negligible thin walls of the same surface area,are filled with water at $90^{\circ}C$ and allowed to cool in the same environment. The graph that correctly represents their cooling is:

$A$ cup of tea cools from $80\,^{\circ}C$ to $60\,^{\circ}C$ in $1\,min$. The ambient temperature is $30\,^{\circ}C$. In cooling from $60\,^{\circ}C$ to $50\,^{\circ}C$,it will take ....... $sec$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module