Question

15.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 70.0°C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings. (The specific heat capacity of H2O(l), cs = 4.18 J/gC; H2O(s)  H2O(l) H = 6.02 kJ/mol)

Answer 1

Answer : The final temperature will be $66.7^oC$

Explanation :

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

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

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 = 15.0 g

$m_2$ = mass of water = 150 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $0.0^oC$

$T_2$ = initial temperature of water = $70.0^oC$

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

$15.0g\times 2.09J/g^oC\times (T_f-0.0)^oC=-150g\times 4.18J/g^oC\times (T_f-70.0)^oC$

$T_f=66.7^oC$

Therefore, the final temperature will be $66.7^oC$