Two samples- each containing 400 g of water- are mixed- The temperature of sample A before mixing was 65- C and that of sample B was 20- C- What will be the final temperature of the mixture- Specific heat of water - 4186 J-kg-K

Given: Mass of sample A and B, m_A=m_B=400 g=0.4 kg Temperature of sample A, T_A=65^∘ C Temperature of sample B, T_B=20^∘ C Specific heat of water, S_w=4186 J/( ...

You visited us 11 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Method of Mixtures

Two samples, ...

Question

Two samples, each containing $400 g$ of water, are mixed. The temperature of sample A before mixing was $65^{\circ} C$ and that of sample B was $20^{\circ} C$ . What will be the final temperature of the mixture? (Specific heat of water = 4186 J/(kg.K)

{40.5}^{\circ} C

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

{41.5}^{\circ} C

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

{42.5}^{\circ} C

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

{43.5}^{\circ} C

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

Open in App

Solution

The correct option is C ${42.5}^{\circ} C$
Given:
Mass of sample A and B, $m_{A} = m_{B} = 400 g = 0.4 k g$
Temperature of sample A, $T_{A} = 65^{\circ} C$
Temperature of sample B, $T_{B} = 20^{\circ} C$
Specific heat of water, $S_{w} = 4186 J / (k g . K)$
The heat lost by sample A will be equal to heat gained by sample B.
Let the temperature of the mixture be $T_{m}$ .
Heat lost = Heat gained
$∴$ $m_{A} S_{w} (T_{A} - T_{m}) = m_{B} S_{w} (T_{B} - T_{)}$
$\Rightarrow 0.4 \times (65 - T_{m}) = 0.4 \times (T_{m} - 20)$
$\Rightarrow 0.8 T_{m} = 34$
$\Rightarrow T_{m} = {42.5}^{\circ} C$

Suggest Corrections

Similar questions

Q. Two samples, each containing

400 g

of water, are mixed. The temperature of sample A before mixing was

20^{\circ} C

and that of sample B was

65^{\circ} C

. What will be the final temperature of the mixture? (Specific heat of water = 4186 J/kg K)

Q. Sample A containing

100 g

of water at

30^{\circ} C

is mixed with another sample B containing

100 g

water at

10^{\circ} C

. What will be the temperature (T) of the mixture of sample A and B?

1 K g

of ice at

- 20^{\circ} C

is mixed with

2 K g

of water at

90^{\circ} C

. Assuming that there is no loss of energy to the environment, what will be the final temperature of the mixture? (Assume latent heat of ice

= 334.4 K J / K g

, specific heat of water and ice are

4.18 k J / (k g . K)

and

2.09 k J / (k g . K)

, respectively.)

Q. A 50 g sample of metal was heated to

100^{0} C

and then dropped into a beaker containing

50 g

of water at

25^{0} C

. If the specific heat capacity of the metal is 0.25 cal/g, what is the final temperature of the water in

^{0} C

Q. What would be the final temperature of the mixture when

5

g of ice at

- 10^{o}

C are mixed with

20

g of water at

30^{o}

C. Specific heat of ice is

0.5

and latent heat of water

80

cal

g^{- 1}

Join BYJU'S Learning Program

Two samples, each containing 400 g of water, are mixed. The temperature of sample A before mixing was 65∘C and that of sample B was 20∘C. What will be the final temperature of the mixture? (Specific heat of water = 4186 J/(kg.K)

Two samples, each containing $400 g$ of water, are mixed. The temperature of sample A before mixing was $65^{\circ} C$ and that of sample B was $20^{\circ} C$ . What will be the final temperature of the mixture? (Specific heat of water = 4186 J/(kg.K)