Question

When 0.2 kg of a body at 100°c id dropped into 0.5 kg of water at 10°c , the resulting temperature is 16°c .
Find the specific heat of the body.
Specific heat of water is 4.2×10^3 j/kg/°c
​

Answer 1

Answer:

please mark me as brain list

given :

time taken by mahesh to reach the maket from his house on his cycle is 15 minutes.

the speed of the cycle is 4 m/s.

\;\large\underbrace{\underline{\sf{understanding\;the\;concept}}}

understandingtheconcept

Answer 2

$\huge\rm\underline\purple{Question :-}$

When 0.2 kg of a body at 100°C is dropped into 0.5 kg of water at 10°C , the resulting temperature is 16°C . Find the specific heat of the body. Specific heat of water is 4.2×10^3 J/kg/°C

$\large{\underbrace{\sf{\red{ Required\:Solution:}}}}$

${\large{\bold{\sf{\bf {\underline{For\: the\: body,}}}}}}$

• m1 = 0.2 kg
• ΔT1 = 100° -16° = 84°C
• s1 = ❓

${\large{\bold{\sf{\bf {\underline{For\: the\: water,}}}}}}$

• m2 = 0.5 kg
• ΔT2 = 16° -10° = 6°C
• s2 = 4.2 × 10^3 J/kg/°C

$\green{\underline\bold{We\:know,}}$

$\boxed{ \sf \orange{ From\: the \: law \: of\: conservation \: of\: energy, }}$

$\boxed{ \sf \purple{ Heat\: lost \: by\: body=\: Heat \: gained\: by\: water}}$

✏ m1 s1 ΔT1 = m2 s2 ΔT2

✏ s1 = $\frac{m2\: s2\: \Delta\: T2}{m1\: \Delta\:T1}$

✏ s1 = $\frac{0.5×(4.2×10^3)×6}{0.2×84}$

✏ s1 = 0.75 × 10^3 J/kg/°C

$\pink{\underline\bold{ ∴ \:The\:required\:answer\:is\:0.75 × 10^3 J/kg/°C}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━