When 40 g of ice at 0-C is added to 160 g of water at 40-C- find the final temperature-

Heat lost by 160g of water at 40 #8451; = 160× 40× 1 =6400 cal Heat required to convert 0 #8451; ice to 0 #8451; water= 40× ...

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

Specific Heat Capacity

When 40 g of ...

Question

When 40 g of ice at $0^{\circ} C$ is added to 160 g of water at $40^{\circ} C$ , find the final temperature?

Open in App

Solution

Suggest Corrections

Similar questions

300

25^{\circ} C

100 g

0^{\circ} C

- 20^{\circ} C

50 g

40^{\circ} C

0^{\circ} C

20 g

(Specific heat of water = 4.2 {Jg}^{- 1}^{\circ} C^{- 1} Specific heat of Ice = 2.1 {Jg}^{- 1}^{\circ} C^{- 1} Heat of fusion of water at 0^{\circ} C = 334 {Jg}^{- 1})

40 g

0^{o} C

200 g

50^{\circ} C

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

336 \times 10^{3} J k g^{- 1}

10

40

15

^{o}

1

0^{o}

5

10^{o}

80 c a l / g

Join BYJU'S Learning Program