A copper block of mass $1 k g$ slides down on a rough inclined plane of inclination $37^{\circ}$ at a constant speed. Find the increase in the temperature of the block as it slides down through $60 c m$ assuming that the loss in mechanical energy goes into the copper block as thermal energy. (specific heat of copper $420 J K g^{- 1} K^{- 1}$ .

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Solution

Specific heat of copper $420 J K g^{- 1} K^{- 1}$

Heat energy absorbed:

Heat energy absorbed:

$Q = m c Δ T . . . . (1)$
Work done is equal to the loss in mechanical energy

$Δ E = W$

$= (m g sin θ) L . . . . . . . . (2)$

From (1) and (2)
$m c Δ T = m g sin θ L$
$Δ T = \frac{g sin θ L}{c}$
$Δ T = \frac{10 sin (37) \times 0.6}{420}$
$Δ T = 8.5 \times 10^{- 3}^{0} C$

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Q. A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37° at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = 420 J kg⁻¹ K⁻¹.

A copper cube of mass $1 k g$ slides down on a rough inclined plane of inclination $37^{\circ}$ at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increases in the temperature of the block as it slides down through $60 c m$ . Specific heat capacity of copper $C_{c}$ = $420 J k g^{- 1} K^{- 1}$ . Take $s i n 37^{\circ}$ = $\frac{3}{5}$

A copper cube of mass 200g slides down on a rough inclined plane of inclination $37^{\circ}$ at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increases in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = $420 J k g^{- 1} K^{- 1}$ .

Q. A copper cube of mass

200 g

slides down on a rough inclined plane of inclination

37^{\circ}

at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides

60 c m

along the inclined plane. Specific heat capacity of copper =

420 J k g^{- 1} K^{- 1}

. Take

s i n (37^{\circ}) = \frac{3}{5}

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Specific heat of copper 420JKg−1K−1 Heat energy absorbed: Heat energy absorbed: Q=mcΔT....(1)Work done is equal to the loss in mechanical energy ΔE=W =(mgsinθ)L........(2) From (1) and (2)mcΔT=mgsinθLΔT=gsinθLcΔT=10sin(37)×0.6420ΔT=8.5×10−30C