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

Effects of Heat

0.120 kg of c...

Question

0.120 kg of copper of specific heat capacity $100 c a l k g^{- 1}^{o} C^{- 1}$ and at $20^{o} C$ is heated by a burner for two minutes, which supplies heat energy at a rate of 20 calories/sec. Calculate the final temperature of copper in ( $C^{o}$ )

Open in App

Solution

$L e t r i s e i n t e m p e r a t u r e = θ_{R}$

$H e a t s u p p l i e d t o c o p p e r = \frac{20 c a l}{s} \times 120 s = 2400 c a l$
Also, $H e a t s u p p l i e d = m c θ_{R}$

$2400 c a l = 0.12 k g \times 100 c a l k g^{- 1}^{o} C^{- 1} θ_{R}$

$∴ θ_{R} = \frac{2400}{0.12 \times 100}^{o} C = 200^{o} C$

The final temperature of the copper
$= 20^{o} C + 200^{o} C$
$= 220^{o} C$ .

Suggest Corrections

Similar questions

0.120

100 \frac{c a l}{K g^{o} C}

20^{o}

20

100 g

20^{o} C

70^{o} C

= 390 J k g^{- 1} K^{- 1}

60 g

100^{o} C

20^{o} C

25^{o} C

k g^{- 1}^{o} C^{- 1}

0.047 k g

100^{\circ} C

0.14 k g

0.25 k g

20^{o} C

23^{o} C

= 4.18 \times 10^{3} J k g^{- 1}^{o} C^{- 1}

0.386 \times 10^{3} J k g^{- 1}^{o} C^{- 1}

200 g

200^{o} C

20^{o} C

0.4 J / g^{o} C

i n J

Join BYJU'S Learning Program