You visited us 4 times! Enjoying our articles? Unlock Full Access!

1st Law of Thermodynamics

How much heat...

Question

How much heat energy is required to prepare a water bath of $20 k g$ , when cold water at $10^{o} C$ is heated to $40^{o} C$ ? (Specific heat capacity of water is $4200 J / k g^{o} C$ ) :

252000000 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

252000 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25200 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2520000 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $2520000 J$
Temperature raise is $40 - 10 = 30^{o} C$
heat required $= m s Δ T$
$= 20 \times 4200 \times 30 = 2520000 J$

Suggest Corrections

Similar questions

75 g

10^{o} C

25200 J

4.2 J / g^{o} C

2 K g

0^{0} C

20^{0} C

3, 34, 000 J / K g

4200 J {K g}^{- 1} K^{- 1}

4200 J / k g^{o} C

10 k g

^{o} C

1 k g

- 10^{o} C

100^{o} C

7, 77, 000 J

= 2100 J k g^{- 1} K^{- 1}

= 4200 J k g^{- 1} K^{- 1}

10 k g

0^{o} C

50^{o} C

4200 J / k g^{o} C

336 \times 10^{3} J / k g

Join BYJU'S Learning Program