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4 years ago

11

A 300 g glass thermometer initially at 32 ◦C is put into 157 cm3 of hot water at 95 ◦C. Find the final temperature of the thermo

meter, assuming no heat flows to the surroundings. The specific heat of glass is 0.2 cal/g · ◦ C and of water 1 cal/g · ◦ C.

1 answer:

Annette [7]4 years ago

6 0

Answer:

T_f= 77.58° C

Explanation:

from simple calorimetry we can write that

$Q_w = m_wc_w(T_f-T_w)$

and

$Q_g = m_gc_g(T_f-T_w)$

Where

Q_w = heat content of water

Q_g= heat content of glass

m_g= mass of glass

m_w= mass of water

T_f= final temp

T_w= temp of water

T_g= temp of glass

m_w =mass of water

m_g= mass of glass

The specific heat of glass is 0.2 cal/g · ◦ C and of water 1 cal/g · ◦ C.

Now in given case

Q_w+Q_g=0

therefore

$Q_w = m_wc_w(T_f-T_w)$ + $Q_g = m_gc_g(T_f-T_w)$ =0

⇒T_f= $\frac{m_gc_gT_g+m_wc_wT_g}{m_wc_w+m_gc_g}$

putting values we get

T_f= $\frac{300\times0.2\times32+157\times1\times95}{157\times1+300\times0.2}$

T_f= 77.58° C

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You solve the equation (5) for v2:

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The final speed of the box is 2.58m/s

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