Question

pls solve this one fast.......​

Answer 1

So we have the equation

$\displaystyle\longrightarrow\sf {m_1(T_1-T_0)=m_2(T_0-T_2)\quad\quad\dots(1)}$

since the two liquids are of same kind or are of having same specific heat capacity, where,

• $\displaystyle\sf {m_1=}$ mass of the hotter liquid

• $\displaystyle\sf {m_2=}$ mass of the colder liquid

• $\displaystyle\sf {T_0=}$ resulting temperature

• $\displaystyle\sf {T_1=}$ temperature of the hotter one

• $\displaystyle\sf {T_2=}$ temperature of the colder one

But here,

$\displaystyle\leadsto\sf {m_2=3m_1}$

$\displaystyle\leadsto\sf {T_0=20^{\circ}}$

$\displaystyle\leadsto\sf {T_2=10^{\circ}}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf {m_1(T_1-20^{\circ})=3m_1(20^{\circ}-10^{\circ})}$

$\displaystyle\longrightarrow\sf {T_1-20^{\circ}=3\times 10^{\circ}}$

$\displaystyle\longrightarrow\sf {T_1=20^{\circ}+30^{\circ}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {T_1=50^{\circ}}}}$

Answer 2

Answer:

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