Question

100 ml of water at 30° C mixed with 200ml of water at 70° C then temperature of mixture is
A) 40° C
B) 56.7° C
C) 70° C
D) 30° C
​

Answer 1

Answer:

Option D: 30°C

Explanation:

Temperature = 100+200/70+30

= 300/100

= 30°C

Answer 2

Answer:

It is accurate that the ultimate temperature was 56.7°C.

Explanation:

Data provided:

Liquid – 1

100 ml of water has a mass of one kilogramme (m1).

Water temperature $(T1) = 30 \textdegree C$, or $(30 + 273) K$, or $303 K$.

Water's c1 specific heat is $4.187 KJ/kgK$.

Liquid - 2

Water mass (m2) is equal to 200 ml or 0.2 kilogramme.

Water's temperature (T2) is 60 °C, or $(273 + 70) K$, or $343 K$.

Water's c2 specific heat is $4.187 KJ/kgK$.

After the two liquids are combined, heat loss by one liquid is offset by heat gain by another, and the two liquids reach a common ultimate temperature.

Let T represent the average final temperature.

$m_{1} \times c_{1} \times (303 - T) = m_{2} \times c_{2} \times ( T - 343 )$

$0.1 \times 4.187 \times ( 303 - T ) = 0.2 \times 4.187 \times ( T - 343 ) ( T - 343 )$

$303 - T = 2T - 686$

$3T = 989$

$T = 329.7 K$

$T = 329.7 - 273 = 56.7\textdegree C$

So the final temperature is 56.7°C

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