149619 An object cools down from $90^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in 10 minutes, when room temperature is $30^{\circ} \mathrm{C}$. The time taken by the object to cool from $70{ }^{0} \mathrm{C}$ to $50{ }^{\circ} \mathrm{C}$ is

149620 A body cools down from $75^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in time $\left(\Delta t_{1}\right)$, from $70{ }^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in time $\left(\Delta t_{2}\right)$ and 65 ${ }^{\circ} \mathrm{C}$ to $60{ }^{\circ} \mathrm{C}$ in time $\left(\Delta \mathrm{t}_{3}\right)$. The correct statement according to Newton's law of cooling is:

149621 A body cools from $70^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in $5 \mathrm{~min}$. Calculate the time it takes to cool from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The temperature of the surrounding is $20^{\circ} \mathrm{C}$.

149622 A metal ball of mass $1 \mathrm{~kg}$ is heated using a 40 $W$ heater in a room at $30^{\circ} \mathrm{C}$. The temperature of the ball becomes steady at $70^{\circ} \mathrm{C}$. Assuming Newton's law of cooling, the rate of loss of heat to the surrounding when the ball is at $40^{\circ} \mathrm{C}$ is

149619 An object cools down from $90^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in 10 minutes, when room temperature is $30^{\circ} \mathrm{C}$. The time taken by the object to cool from $70{ }^{0} \mathrm{C}$ to $50{ }^{\circ} \mathrm{C}$ is

149620 A body cools down from $75^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in time $\left(\Delta t_{1}\right)$, from $70{ }^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in time $\left(\Delta t_{2}\right)$ and 65 ${ }^{\circ} \mathrm{C}$ to $60{ }^{\circ} \mathrm{C}$ in time $\left(\Delta \mathrm{t}_{3}\right)$. The correct statement according to Newton's law of cooling is:

149621 A body cools from $70^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in $5 \mathrm{~min}$. Calculate the time it takes to cool from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The temperature of the surrounding is $20^{\circ} \mathrm{C}$.

149622 A metal ball of mass $1 \mathrm{~kg}$ is heated using a 40 $W$ heater in a room at $30^{\circ} \mathrm{C}$. The temperature of the ball becomes steady at $70^{\circ} \mathrm{C}$. Assuming Newton's law of cooling, the rate of loss of heat to the surrounding when the ball is at $40^{\circ} \mathrm{C}$ is

149619 An object cools down from $90^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in 10 minutes, when room temperature is $30^{\circ} \mathrm{C}$. The time taken by the object to cool from $70{ }^{0} \mathrm{C}$ to $50{ }^{\circ} \mathrm{C}$ is

149620 A body cools down from $75^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in time $\left(\Delta t_{1}\right)$, from $70{ }^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in time $\left(\Delta t_{2}\right)$ and 65 ${ }^{\circ} \mathrm{C}$ to $60{ }^{\circ} \mathrm{C}$ in time $\left(\Delta \mathrm{t}_{3}\right)$. The correct statement according to Newton's law of cooling is:

149621 A body cools from $70^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in $5 \mathrm{~min}$. Calculate the time it takes to cool from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The temperature of the surrounding is $20^{\circ} \mathrm{C}$.

149622 A metal ball of mass $1 \mathrm{~kg}$ is heated using a 40 $W$ heater in a room at $30^{\circ} \mathrm{C}$. The temperature of the ball becomes steady at $70^{\circ} \mathrm{C}$. Assuming Newton's law of cooling, the rate of loss of heat to the surrounding when the ball is at $40^{\circ} \mathrm{C}$ is

149619 An object cools down from $90^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in 10 minutes, when room temperature is $30^{\circ} \mathrm{C}$. The time taken by the object to cool from $70{ }^{0} \mathrm{C}$ to $50{ }^{\circ} \mathrm{C}$ is

149620 A body cools down from $75^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in time $\left(\Delta t_{1}\right)$, from $70{ }^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in time $\left(\Delta t_{2}\right)$ and 65 ${ }^{\circ} \mathrm{C}$ to $60{ }^{\circ} \mathrm{C}$ in time $\left(\Delta \mathrm{t}_{3}\right)$. The correct statement according to Newton's law of cooling is:

149621 A body cools from $70^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in $5 \mathrm{~min}$. Calculate the time it takes to cool from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The temperature of the surrounding is $20^{\circ} \mathrm{C}$.

149622 A metal ball of mass $1 \mathrm{~kg}$ is heated using a 40 $W$ heater in a room at $30^{\circ} \mathrm{C}$. The temperature of the ball becomes steady at $70^{\circ} \mathrm{C}$. Assuming Newton's law of cooling, the rate of loss of heat to the surrounding when the ball is at $40^{\circ} \mathrm{C}$ is