Question

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam container so that heat loss can be ignored. Find the final temperature.

Answer 1

Answer : The final temperature is,
`25.0^oC`

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of ice =
`2.09J/g^oC`

`c_2` = specific heat of water =
`4.18J/g^oC`

`m_1` = mass of ice = 50 g

`m_2` = mass of water = 200 g

`T_f` = final temperature = ?

`T_1` = initial temperature of ice =
`-15^oC`

`T_2` = initial temperature of water =
`30^oC`

Now put all the given values in the above formula, we get:

`50g* 2.09J/g^oC* (T_f-(-15))^oC=-200g* 4.184J/g^oC* (T_f-30)^oC`

`T_f=25.0^oC`

Therefore, the final temperature is,
`25.0^oC`