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Question
A metal block of heat capacity 80 J°C−1 placed in a room at 20°C is heated electrically. The heater is switched off when the temperature reaches 30°C. The temperature of the block rises at the rate of 2°C s−1 just after the heater is switched on and falls at the rate of 0.2°C s−1 just after the heater is switched off. Assume Newton's law of cooling to hold.
- Find the power of the heater.
- Find the power radiated by the block just after the heater is switched off.
- Find the power radiated by the block when the temperature of the block is 25°C.
- Assuming that the power radiated at 25°C represents the average value in the heating process, find the time for which the heater was kept on.
Solution
Given:
Heat capacity of the metal block, s = 80 J°C−1
Heat absorb by the metal block is,
`H = s xx DeltaT = 80xx(30-10) = 800 J`
Rate of rise in tempreature of the block `= ((d theta)/(dt))=2^circC//s`
Rate of fall in temperature of the block = `((d theta)/(dt)) = - 0.2^circ C//s`
The negative sign indicates that the temperature is falling with time.
a. Energy = s `(Delta theta)`
(Power = Energy per unit time
∴ Power of the heater = Heat capacity ×`(d theta)/(dt)`
P = 80 × 2
P = 160 w
b. Power Radiated = Energy lost per unit time.
∴ P = heat capacity ×`((d theta)/(dt))`
Here, `(d theta)/(dt)` represents the rate of decrease in temperature.
p = 80 × 0.2
p = 16 W
c. 16 W of power is radiated when the temperature of the block decreases from 30°C to 20°C.
∴ `s xx ((d theta)/(dt))_{dec} = k (theta - theta_0)`
⇒ 16 = K (30 − 20)
K = 1.6
From newton's law of cooling,
∴ s`(d theta)/(dt) = K (theta - theta_0)`
`⇒ s(d theta)/(dt) = 1.6 (30 - 25)`
`⇒ s(d theta)/(dt) = 1.6 xx 5`
`⇒ s(d theta)/(dt) = 8 "Watt"`
d. As we know, that power given by the heater is P = 160 W
now we can use energy equations
Energy given by heater = Heat absorbed by block + heat loss
160t = 80(30 − 20) + 8t
154t = 800
t = 5.19 s
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