Question

4 grams of liquid a at 60 degree celsius is mixed with one gram of liquid B at 50 degree Celsius if final temperature is 55 degree Celsius then find the ratio of the specific heats​

Answer 1

Answer:

Kisi Unit mass Ka Temperature , Unit Kelvin badane ke Liye Heat = Specific Heat

Agar m mass Ka Temperature T Kelvin Badana then

Heat = mS∆T

S = Q/m∆T

∆T = Change In Temperature

∆T = Tf - Ti

I sent a Attachment

Ratio -

1:4

Answer 2

The ratio of the specific heat is 1 : 4

Explanation:

Given:

• 4 grams of liquid a at 60 degree celsius
• 1 gram of liquid B at 50 degree Celsius
• Final temperature is 55 degree

To find: The ratio of the specific heat

Solution:

Liquid A

$m_{A} = 4 gms$

Angle = 60 C

Liquid B

$m_{B} = 1 gm$

Angle = 50 C

$O_{R} = \frac{m_{A} S_{A} O_{A} +m_{B} S_{B} O_{B}}{m_{A} S_{A} +m_{B} S_{B} }$

Specific heat = $\frac{S_{A} (4)(60) + S_{B} (1)(50)}{4S_{A} + 1 .S_{B} }$

Specific heat = $\frac{S_{B} [\frac{S_{A} }{S_{B} }240+50]}{S_{B} [\frac{S_{A} }{S_{B} }4+1]}$

⇒ $220(\frac{S_{A} }{S_{B} } )+55= 240(\frac{S_{A} }{S_{B} } )+50$

⇒ $\frac{S_{A} }{S_{B} } )(240 - 220) = 55 - 50$

⇒ $\frac{S_{A} }{S_{B} } = \frac{5}{20}$

⇒ $\frac{S_{A} }{S_{B} } = \frac{1}{4}$

$S_{A} :S_{B} = 1:4$

Final answer:

The ratio of the specific heat is 1 : 4

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