A liquid at 30°C is poured very slowly into a calorimeter that is at temperature of 110°C. The boiling temperature of the liquid is 80°C. It is found that the first 5 gm of the liquid completely evaporates. After pouring another 80 gm of the liquid, its specific heat will be ____ °C.

[Neglect the heat exchange with surrounding] [2019]

Question

A liquid at 30°C is poured very slowly into a calorimeter that is at temperature of 110°C. The boiling temperature of the liquid is 80°C. It is found that the first 5 gm of the liquid completely evaporates. After pouring another 80 gm of the liquid, its specific heat will be ____ °C.

[Neglect the heat exchange with surrounding] [2019]

Smart Achievers · Accepted Answer

(270)

$Let C be the specific heat capacity of liquid and L be the latent heat of vapourisation.$

$From principle of calorimetry,$

$Heat lost = heat gain$

$m_{c} S_{c} Δ T = m C Δ T + m L$

$or m_{c} S_{c} (110 - 80) = 5 C (80 - 30) + 5 L$ ...(i)

$Where, m_{c} = mass of calorimeter$

$S_{c} = sp. heat of calorimeter$

$Again, when 80 g liquid is poured and equilibrium temperature is 50 ° C$

$m_{c} S_{c} (80 - 50) = 80 C (50 - 30)$ ...(ii)

$From eq. (i) & (ii)$

$1600 C = 250 C + 5 L$

$∴$ $\frac{L}{C} = \frac{1350}{5} = 270 ° C$

Solution

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