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A piece of glass with a mass of 32.50 g specific heat of 0.840 J/g°C and an initial temperature of 115 °C was dropped into a calorimeter containing 57 g of water (specific heat 4.184 J/g°C). The final temperature of the glass and water in the calorimeter was 119.2 °C. What was the initial temperature of the water?



39.84°C

79.68°C

119.84°C

139.68°C

Verified Answer :

Answer : The final temperature of water is, \(119.84^oC\)

Solution :

\(Q_{absorbed}=Q_{released}\)

As we know that,  

\(Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})\)

\(m_1\times c_g\times (T_{final}-T_2)=-[m_2\times c_w\times (T_{final}-T_1)]\)          .................(1)

where,

\(m_1\) = mass of glass = 32.50 g

\(m_2\) = mass of water = 57 g

\(T_{final}\) = final temperature  of water and glass = \(119.2^oC\)

\(T_2\) = initial temperature of water = ?

\(T_1\) = initial temperature glass = \(115^oC\)

\(c_w\) = specific heat of water = \(4.184J/g^oC\)

\(c_g\) = specific heat of glass = \(0.840J/g^oC\)

Now put all the given values in equation (1), we get

\(m_1\times c_g\times (T_{final}-T_2)=-[m_2\times c_w\times (T_{final}-T_1)]\)

\((32.50g)\times (0.840J/g^oC)\times (119.2^oC-115^oC)=-[(57g)\times (4.184J/g^oC)\times (119.2^oC-T_1)]\)

\(T_1=119.84^oC\)

Therefore, the final temperature of water is, \(119.84^oC\)

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