2 kg water at 80°C is mixed with 3 kg water at 20°C. Assuming no heat losses, find the final temperature of
the mixture.
Answers
Explanation:
Answer:
The Final temperature of the mixture (T) is 44°C
Given:
Mass of Water (M₁) = 2 Kg.
Temperature of Water (T₁) = 80°C
Mass of Water (M₂) = 3 Kg.
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}⟼
HeatLostby2KgWater
M
1
CΔT
1
=
HeatGainedby3KgWater
M
2
CΔT
2
Substituting the Values,
\longmapsto \large{\tt 2 \; Kg \times C \times (80 - T) = 3 \; Kg \times C \times (T - 20)}⟼2Kg×C×(80−T)=3Kg×C×(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)}⟼2Kg×
C
×(80−T)=3Kg×
C
×(T−20)
Simplifying,
\longmapsto \large{\tt 2 \times (80 - T) = 3 \times (T - 20)}⟼2×(80−T)=3×(T−20)
\longmapsto \large{\tt 160 - 2 \; T = 3 \; T - 60}⟼160−2T=3T−60
\longmapsto \large{\tt 160 + 60 = 3 \; T + 2 \; T}⟼160+60=3T+2T
\longmapsto \large{\tt 220 = 5 \; T}⟼220=5T
\longmapsto \large{\tt 5 \; T = 220}⟼5T=220
\longmapsto \large{\tt T = \dfrac{220}{5}}⟼T=
5
220
\longmapsto \large{\tt T = \cancel{\dfrac{220}{5}}}⟼T=
5
220
\longmapsto \large{\underline{\boxed{\red{\tt T = 44 ^\circ C}}}}⟼
T=44
∘
C
∴ The Final temperature of the mixture (T) is 44°C.