Question

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

Answer 1

119.84 c is the answer

Answer 2

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, $119.84^oC$