Question

A metal cylinder of mass 0.5 kg is heated electrically by 12 W heater in a room at 15°c. The cylinder temperature rise uniformly to 25°c in 5 min and later becomes constant at 45°c. Calculate the specific heat capacity of the metal taking into account the loss of heat to the surrendering ?

Answer: 600J/kg.K​

Answer 1

Answer:

The required answer is $2$ Watt.

Step-by-step explanation:

Given: A metal cylinder with a mass of $0.5 kg$ is electrically heated in a $15$°C chamber by a $12 W$ heater. The temperature in the cylinder rises consistently to $25$°C in $5$ minutes before stabilizing at $45$°C

To find: the specific heat capacity of the metal taking into account the loss of heat to the surrendering.

According to question,

The heat loss from the cylinder will equal the heat gain by the heater at steady state.

Room temperature is $\mathrm{T}_{0}=15^{\circ} \mathrm{C}$

We must take into account that the material enters steady state at $45^{\circ} \mathrm{C}$ in order to determine its emissivity.

so heat loss $\mathrm{H}=\mathrm{K}\left(\mathrm{T}-\mathrm{T}_{0}\right)$

A proportionality constant is $\mathrm{K}$ and the heat loss rate is $\mathrm{H}$

At $45^{\circ} \mathrm{C}$ it will be $\mathrm{K}(45-15)=30 \mathrm{~K}$ watt

It should be equal to $12 \mathrm{~W}$ so $\mathrm{K}=\frac{12}{30}=\frac{2}{5}$

Now for $20^{\circ} \mathrm{c}$

Hence, The rate of heat loss will be $E=K(20-15)=5 \mathrm{~K}=2 \mathrm{~W}$att.

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