Name: Molecular Arrangement in Gases
Uploaded: 2019-07-19
Duration: 5 min 44 s
Description: Do you know why gases have no fixed shape, size, and volume? It is because the molecules of a gas are far apart and constantly colliding with each other. Wat...

Question

A calorimeter of mass

90 g

and specific heat capacity

420 J {kg}^{- 1} K^{- 1}

contains

310 g

of a liquid at

15.0 ° C

. An electric heater rated at

20.0 W

warms the liquid to

19.0 ° C

in

3.0 \min

. Assuming there are no energy losses to the surroundings, estimate the specific heat capacity of the liquid.

Accepted Answer

Given,

Mass of calorimeter, $M = 90 g = 0.090 kg$

Mass of liquid, $m = 310 g = 0.310 kg$

Specific heat capacity of the material of the calorimeter, $c_{1} = 420 J {kg}^{- 1} K^{- 1}$

The initial temperature of the calorimeter and liquid, $T_{1} = 15 ° C$

The final temperature of the calorimeter and liquid, $T_{2} = 19 ° C$

$∴$ The change in temperature, $∆ T = T_{2} - T_{1} = 19 ° C - 15 ° C = 4 ° C$

Power of electric heater, $P = 20.0 W$

Time, $t = 3.0 \min$

The energy provided by the electric heater is, $E = P \times t = 20.0 \times 3.0 \times 60 = 3600 J$

Let $c_{2}$ be the specific heat capacity of the liquid.

Assuming no loss of heat to the surrounding, the heat lost by the electric heater goes into the calorimeter and liquid. Thus,

$M \times c_{1} \times ∆ T + m \times c_{2} \times ∆ T = 3600$ $\to e q u a t i o n (1)$

$($ The temperature changes between kelvin and $° C$ are the same in both scales. $)$

Substituting all the values in $e q u a t i o n (1)$ , we get,

$0.090 \times 420 \times 4 + 0.310 \times c_{2} \times 4 = 3600$

$\Rightarrow c_{2} = \frac{3600 - 151.2}{1.24} = 2781.29 J {kg}^{- 1} K^{- 1}$

So the specific heat capacity of the liquid is $2781.29 J {kg}^{- 1} K^{- 1}$ .

Important Questions on Thermal Physics

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A calorimeter of mass 90 g and specific heat capacity 420 J kg-1 K-1 contains 310 g of a liquid at 15.0°C. An electric heater rated at 20.0 W warms the liquid to 19.0°C in 3.0 min. Assuming there are no energy losses to the surroundings, estimate the specific heat capacity of the liquid.

Important Questions on Thermal Physics

Learn and Prepare for any exam you want !