SATHEE: Heat and Thermodynamics 1 Question 5

Heat and Thermodynamics 1 Question 5

5. A metal ball of mass $0.1 k g$ is heated upto $500^{\circ} C$ and dropped into a vessel of heat capacity $800 J K^{- 1}$ and containing 0.5 $k g$ water. The initial temperature of water and vessel is $30^{\circ} C$ . What is the approximate percentage increment in the temperature of the water? [Take, specific heat capacities of water and metal are respectively $4200 J k g^{- 1} K^{- 1}$ and $400 J k g^{- 1} K^{- 1}$ ]

(2019 Main, 11 Jan II)

(a) $25$

(b) $15$

(d) $20$

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Solution:

Using heat lost or gained without change in state is $Δ Q = m s Δ T$ , where $s$ is specific heat capacity and $T =$ change in temperature

Let final temperature of ball be $T$ .

Then heat lost by ball is,

$Δ Q = 0.1 \times 400 (500 - T)$

This lost heat by ball is gained by water and vessel and given as

Heat gained by water,

$Δ Q_{1} = 0.5 \times 4200 (T - 30)$

and heat gained by vessel is

$\begin{aligned} Δ Q_{2} & = heat capacity \times Δ T \\ = 800 \times (T - 30) \end{aligned}$

According to principle of calorimetry, total heat lost $=$ total heat gained

$\begin{aligned} \Rightarrow & 0.1 \times 400 (500 - T) \\ = & 0.5 \times 4200 (T - 30) + 800 (T - 30) \end{aligned}$

$\begin{array}{ll} \Rightarrow & (500 - T) = \frac{(2100 + 800) (T - 30)}{40} \\ \Rightarrow & 500 - T = 72.5 (T - 30) \\ \Rightarrow & 500 + 217.5 = 72.5 T or T = 36.39 K \end{array}$

So, percentage increment in temperature of water

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Heat and Thermodynamics 1 Question 5

Solution:

इंजीनियरिंग की तैयारी

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सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Heat and Thermodynamics 1 Question 5

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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