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

Answer : The final temperature of water is,
`119.84^oC`

Solution :

`Q_(absorbed)=Q_(released)`

As we know that,

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_g* (T_(final)-T_2)=-[m_2* c_w* (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* c_g* (T_(final)-T_2)=-[m_2* c_w* (T_(final)-T_1)]`

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

`T_1=119.84^oC`

Therefore, the final temperature of water is,
`119.84^oC`