## BdPhO Regional (Dhaka-South) Secondary 2019/4

- SINAN EXPERT
**Posts:**38**Joined:**Sat Jan 19, 2019 3:35 pm**Location:**Dhaka, Bangladesh-
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### BdPhO Regional (Dhaka-South) Secondary 2019/4

Consider an isolated system consisted of a ball, and a bucket of water. The ball is released from height, $H$ above a bucket of water. The initial temperature of the water-bucket system and the ball are $T_1$ and $T_2$ respectively. What will be the final temperature of the water after the ball is dropped? (mass of the ball = $m_1$, mass of water = $m_2$, mass of the bucket = $m_3$, specific heat of ball, water and bucket are $s_1$,$s_2$ and $s_3$ respectively)

**$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$**

- SINAN EXPERT
**Posts:**38**Joined:**Sat Jan 19, 2019 3:35 pm**Location:**Dhaka, Bangladesh-
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### Re: BdPhO Regional (Dhaka-South) Secondary 2019/4

As the system is isolated, energy cannot enter or get out from it. The ball is released from height, $H$. When it touches the ground, the kinetic energy converts into sound and thermal energy. I mean,

$Kinetic$ $Energy$ $=$ $Sound$ $Energy$ $+$ $Thermal$ $Energy$

But in fact, the sound energy is negligible. So we can say, $Kinetic$ $Energy$ $=$ $Thermal$ $Energy$

According to the law of calorimetry, LOST HEAT = GAINED HEAT

Let the final temperature of water be $T$.

Hence, $m_1s_1(T_2-T)+m_1gH=(m_2s_2+m_3s_3)(T-T_1)$

$∴T=\dfrac{m_1gH+m_1s_1T_2+(m_2s_2+m_3s_3)T_1}{m_1s_1+m_2s_2+m_3s_3}$

$Kinetic$ $Energy$ $=$ $Sound$ $Energy$ $+$ $Thermal$ $Energy$

But in fact, the sound energy is negligible. So we can say, $Kinetic$ $Energy$ $=$ $Thermal$ $Energy$

According to the law of calorimetry, LOST HEAT = GAINED HEAT

Let the final temperature of water be $T$.

Hence, $m_1s_1(T_2-T)+m_1gH=(m_2s_2+m_3s_3)(T-T_1)$

$∴T=\dfrac{m_1gH+m_1s_1T_2+(m_2s_2+m_3s_3)T_1}{m_1s_1+m_2s_2+m_3s_3}$

**$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$**