Question

Initially a beaker has $100$g of water at temperature ${90}^{o}C$. Later another $600g$ of water at temperature ${20}^{o}C$ was poured into the beaker. The temperature, T of the water after mixing is?

• A. ${20}^{o}C$
• B. ${30}^{o}C$
• C. ${45}^{o}C$
• D. ${55}^{o}C$
• E. ${90}^{o}C$

Answer 1

Answer: (B)

${30}^{o}C$

Given,

Mass of water at ${90}^{o}C=100$gm

$=100\times {10}^{-3}$kg

Mass of water at ${20}^{o}C=600$gm

$=600\times {10}^{-3}$kg

From calorimetery

$m_1{s}_1{t}_1+m_2{s}_2{t}_2=(m_1+m_2)s.T$

$\because s_1{t}_1+s_2{t}_2=st$

[where, T is temperature of mixture].

$100\times {10}^{-3}\times 1\times 90+600\times {10}^{-3}\times 1\times 20$

$=(100+600)\times {10}^{-3}\times 1\times T$

$T=\frac{100\times {10}^{-3}\times 90+600\times {10}^{-3}\times 20}{700\times {10}^{-3}}$

$=\frac{(900+12000)\times {10}^{-3}}{700\times {10}^{-3}}$

$=\frac{21000}{700}$

$={30}^{o}C$.