A calorimeter of negligible heat capacity contains

Question

A calorimeter of negligible heat capacity contains 100 cc of water at 40C. The water cools to 35°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40°C. Find the time taken for the temperature to become 35°C under similar conditions. Specific heat capacities of water and K-oil are 4200 J/kg-K and 2100 J/kg-K respectively. Density of K-oil = 800 kg/m3.

Navjyot Kalra · Accepted Answer

Sol. v = 100 cc ∆θ = °C T = 5 min For water mS∆θ/dt = KA/l ∆θ ⇒ 100 * 10^-3 * 1000 * 4200/5 = KA/l For Kerosene Ms/at = KA/l ⇒ 100 * 10^-3 * 800 * 2100/t = KA/l ⇒ 100 * 10^-3 * 800 * 2100/t = 100 * 10^-3 * 1000 * 4200/5 ⇒ T = 5 * 800 * 2100/1000 * 4200 = 2 min

Thermal Physics> A calorimeter of negligible heat capacity...

Amit Saxena , 11 Years ago

Navjyot Kalra

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Thermal Physics> A calorimeter of negligible heat capacity...

Amit Saxena , 11 Years ago

Navjyot Kalra

LIVE ONLINE CLASSES

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