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Specific Heat

The quantity ...

Question

The quantity of heat which can rise the temperature of x gm of a substance through $t_{1}^{o} C$ can rise the temperature of y gm of water through $t_{2}^{o} C$ is same. The ratio of specific heats of the substances is:

y t_{1} / x t_{2}

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x t_{2} / y t_{1}

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y t_{2} / x t_{1}

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x t_{1} / y t_{2}

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Solution

The correct option is C $y t_{2} / x t_{1}$
Solution
For that substance,
$Δ Q = m s_{1} Δ T = (x) (s_{1}) (t_{1})$ ___ (i)
For water,
$Δ Q = m^{'} s_{2} Δ T = y (s_{2}) (t_{2})$ ___ (ii)
From the $e q^{n}$ (i) and (ii),
$(x) (s_{1}) t_{1} = (y t_{2}) s_{2}$
$\frac{s_{1}}{s_{2}} = \frac{y t_{2}}{x t_{1}} \to$ Water equt
where, $s_{1} \to$ specific heat of substance
$s_{2} \to$ specific heat of water
Option - C is correct

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Similar questions

x g

t_{1}^{o} C

y g

t_{2}^{o} C

The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₂ to t₁?

(a) x(t₂) = x(t₁) + v(t₁)(t₂–t₁) + ()a(t₂–t₁)²

(b) v(t₂)= v(t₁) + a(t₂–t₁)

(d) a_Average = (v(t₂) – v(t₁)) / (t₂ – t₁)

(e) x(t₂) = x(t₁) + v_Average(t₂ – t₁) + ()a_Average(t₂ – t₁)²

(f) x(t₂) – x(t₁) = area under the v–t curve bounded by the t-axis and the dotted line shown.

t (^{\circ} C)

c = 0.20 + 0.14 t + 0.023 t^{2} (c a l / g m^{\circ} C)

5^{\circ} C t o 15^{\circ} C

50

10^{o}

10

= 420

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

= 4200

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

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