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/( ...

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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

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{41.5}^{\circ} C

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{42.5}^{\circ} C

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{43.5}^{\circ} C

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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
$∴$ $m_{A} S_{w} (T_{A} - T_{m}) = m_{B} S_{w} (T_{B} - T_{)}$
$\Rightarrow 0.4 \times (65 - T_{m}) = 0.4 \times (T_{m} - 20)$
$\Rightarrow 0.8 T_{m} = 34$
$\Rightarrow T_{m} = {42.5}^{\circ} C$

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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)