A mass of 50 g of a certain metal at 150-C is immersed in 100 g of water at 11-C- The final temperature is 20-C- Calculate the specific heat capacity of the metal- Assume that the specific heat capacity of water is 4-2 J g-1-K-1-

Heat lost by the metal = Heat gained by water ⇒ m_m S_m Δ T_m = m_w S_w Δ T_w ⇒ 50 × S_m × (150 20) = 100 × 4.2 × (20 11) ⇒ S_m = 0.582 Jg^ 1K^ 1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Principle of Calorimetry

A mass of 50 ...

Question

A mass of 50 g of a certain metal at $150^{\circ} C$ is immersed in 100 g of water at $11^{\circ} C$ . The final temperature is $20^{\circ} C$ . Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 J $g^{- 1} K^{- 1}$ .

Open in App

Solution

Suggest Corrections

Similar questions

50 g

150^{\circ} C

100 g

11^{\circ} C

20^{\circ} C

4.2 J g^{- 1} K^{- 1}

200 g

83^{\circ} C

300 g

30^{\circ} C

33^{o} C

4.2 J g^{- 1} K^{- 1}

150^{\circ} C

11^{\circ} C

20^{\circ} C

4.2 J / g^{\circ} C

100 g

- 10^{\circ} C

100^{\circ} C

2.1 J g^{- 1} K^{- 1}

4.2 J g^{- 1} K^{- 1}

336 J g^{- 1}

45 g

50^{\circ} C

50 g

18^{\circ} C

40.39 J g^{- 1} K^{- 1}

4.2 J g^{- 1} K^{- 1}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program