The figure below shows the variation of speci | vedclass

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

Question

(D) The rate of heat absorption is given by $R = \frac{dq}{dt} = m C \frac{dT}{dt}$. Since the temperature is increased at a constant rate,$\frac{dT}{

Vedclass · Accepted Answer

$(B, C, D)$

Vedclass · Answer

(D) The rate of heat absorption is given by $R = \frac{dq}{dt} = m C \frac{dT}{dt}$. Since the temperature is increased at a constant rate,$\frac{dT}{dt} = k$ (constant). Thus,$R \propto C$.$(A)$ In the range $0-100 \ K$,the graph of $C$ vs $T$ is non-linear (curved),so the rate of heat absorption $R$ does not vary linearly with $T$. Statement $(A)$ is incorrect.$(B)$ The heat absorbed is $\Delta Q = \int m C dT$,which is proportional to the area under the $C-T$ curve. The area under the curve from $0-100 \ K$ is clearly smaller than the area under the curve from $400-500 \ K$ (where $C$ is nearly constant at its maximum value). Statement $(B)$ is correct.$(C)$ In the range $400-500 \ K$,the graph of $C$ vs $T$ is a horizontal line,meaning $C$ is constant. Since $R \propto C$,the rate of heat absorption remains constant. Statement $(C)$ is correct.$(D)$ In the range $200-300 \ K$,the graph of $C$ vs $T$ has a positive slope,meaning $C$ is increasing. Since $R \propto C$,the rate of heat absorption increases. Statement $(D)$ is correct.Therefore,statements $(B, C, D)$ are correct.

Similar Questions

The factor not needed to calculate heat lost or gained when there is no change of state is

The radiator of an engine is cooled by water because:

$100 \ g$ of water is heated from $30^\circ C$ to $50^\circ C$. Ignoring the slight expansion of the water,the change in its internal energy is .......$kJ$ (specific heat of water is $4184 \ J/kg/K$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module