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

Floating and Sinking

The piece of ...

Question

The piece of zinc at a temperature of $20^{o} C$ weighing $65.38 g$ is dropped into $180 g$ of boiling water ( $T = 100^{o} C$ ). The specific heat of zinc is $0.400 J g^{{- 1}^{o}} C^{- 1}$ and that of water is $4.20 J g^{{- 1}^{o}} C^{- 1}$ . What is the final common temperature reached by both the zinc and water :-

{97.3}^{o} C

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

{33.4}^{o} C

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

{80.1}^{o} C

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

60^{o} C

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

Open in App

Solution

The correct option is A ${97.3}^{o} C$
$m_{Z n} . S_{Z n} . (T_{f} - T_{i}) + m_{H_{2} O} . S_{H_{2} O} . (T_{f} - T_{i}) = 0$

$⟹ (65.38 g m) (0.4 J / g^{o} C) (T_{f} - 20^{o} C) + 180 g m (4.20 J / g^{o} C) \times {(T}_{f} - 100^{o} C) = 0$

$⟹ [(65.38) (0.4) + 180 (420)] T_{f} = (65.38) (0.4) (20) + (180) (4.20) (100)$

$T_{f} = [\frac{(65.38) (0.4) (20) + (180) (4.20) (100)}{(65.38) (0.4) + (180) (4.20)}] = {97.3}^{o} C$

Suggest Corrections

Similar questions

20^{o} C

(T = 100^{o} C) .

0.4 J g^{- 1 \circ} C^{- 1}

4.2 J g^{- 1 o} C^{- 1} .

60^{o} C

20^{o} C

100^{o} C

20^{o} C

23^{o} C

100^{o} C

25^{o} C

0.42 J g^{- 1}^{o} C^{- 1}

50 g

0.42 J g^{- 1} ℃^{- 1}

20^{o} C

20 g

100^{o} C

22^{o} C

= 0.3 J g^{- 1} ℃^{- 1}

= 4.2 J g^{- 1} ℃^{- 1}

Join BYJU'S Learning Program