Question

2kg water at 80 degrees Celsius is mixed with 3kg water at 20 degrees Celsius. Assuming no heat losses, find the final temperature of mixture.​

Answer 1

The final temperature is 44°C.

Explanation:

• Mass of water = 2 kg
• Initial temperature = 80° C
• Mass of water = 3 Kg
• Temperature of second water body = 20°

Solution:

The formula is given:

m2 ( T - 20 ) = m1 ( 80 - T )

3 (T - 20) = 2 (80 - T)

3T - 60 = 160 - 2T

3T + 2T = 160 + 60

5 T = 220

T = 220 / 5

T = 44° C

Thus the final temperature is 44°C.

Also learn more

The temperature of 600 gram of cold water rises by 15 degree Celsius when 300 gram of water at 50 degree Celsius is added to it .the initial temperature of the cold water is. ?

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Answer 2

Answer:

• The Final temperature of the mixture (T) is 44°C

Given:

1. Mass of Water (M₁) = 2 Kg.
2. Temperature of Water (T₁) = 80°C
3. Mass of Water (M₂) = 3 Kg.
4. Temperature of Water (T₂) = 20°C

Explanation:

$\rule{300}{1.5}$

As question states that there is no heat loss.

Therefore, we can write,

$\longmapsto \large \tt \underbrace{\tt M_1 \; C \; \Delta T_1}_{Heat \; Lost \; by \; 2\; Kg \; Water} = \tt \underbrace{\tt M_2 \; C \; \Delta T_2}_{Heat \; Gained \; by \; 3\; Kg \; Water}$

Substituting the Values,

$\longmapsto \large{\tt 2 \; Kg \times C \times (80 - T) = 3 \; Kg \times C \times (T - 20)}$

Specific heat will be same as In both conditions Water is Added.

$\longmapsto \large{\tt 2 \; Kg \times \cancel{C} \times (80 - T) = 3 \; Kg \times \cancel{C} \times (T - 20)}$

Simplifying,

$\longmapsto \large{\tt 2 \times (80 - T) = 3 \times (T - 20)}$

$\longmapsto \large{\tt 160 - 2 \; T = 3 \; T - 60}$

$\longmapsto \large{\tt 160 + 60 = 3 \; T + 2 \; T}$

$\longmapsto \large{\tt 220 = 5 \; T}$

$\longmapsto \large{\tt 5 \; T = 220}$

$\longmapsto \large{\tt T = \dfrac{220}{5}}$

$\longmapsto \large{\tt T = \cancel{\dfrac{220}{5}}}$

$\longmapsto \large{\underline{\boxed{\red{\tt T = 44 ^\circ C}}}}$

∴ The Final temperature of the mixture (T) is 44°C.

$\rule{300}{1.5}$