Question

102. Find the temperature of a mixture of 30 gms the of water at 40 °C temperature and 70 gms of
water at 20 °C temperature.

(1) 16°C.
(2)70°C.
(3)24°C.
(4)40°C.​

Answer 1

AnswEr :

• Final temperature of the water mixture (T) is 26°C. [Check your options once]

Step-by-step explanation :

As per the question, 30 gm of water is mixed at temperature of 40°C. Also, 70 gm of water is mixed at temperature of 20°C.

GivEn :

• Mass of the water at 40°C (m1) = 30 gm

• Mass of the water at 20°C (m2) = 70 gm

• Temperature of the water weighing 30 gm (T1) = 40°C

• Temperature of the water weighing 70 gm (T2) = 20°C

To find :

• Final temperature of the water mixture (T) = ?

Formula UsEd :

To calculate the final temperature of the water mixture we know the formula,

$\star \: { \boxed{ \sf{T(final) = \dfrac{(m1 \times {T1} + m2 \times {T2})}{(m1 + m2)} }}}$

where,

• $\sf m_1$ is the weights of the water in the 1st containers .

• $\sf m_2$ the weights of the water in the 2nd containers .

• T1 is the temperature of the water in the 1st container .

• T2 is the temperature of the water in the 2nd container.

SoluTion :

According to the formula,

$\implies \: \: \: \sf{T(final) = \dfrac{(m1 \times T 1+ m2 \times T2)}{(m1 + m2)} }$

Substituting the values,

$\implies \: \: \: \sf{T(final) = \dfrac{(30 \times 40 + 70 \times 20)}{(30 + 70)} }$

$\implies \: \: \: \sf{T(final) = \dfrac{(1200 + 1400)}{(100)} }$

$\implies \: \: \: \sf{T(final) = \dfrac{2600}{100} }$

$\implies \: \: \: \boxed {\sf{T(final) = {26}^{ \circ} }} \: \bigstar$

Therefore, final temperature of the water mixture (T) is 26°C .

Answer 2

Answer:

B.24 is correct answer

Explanation:

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