Question

75.0 kilojoules of energy are added to 125. g of ice at 0.0 ̊C. What is the final temperature of the water?

Answer 1

$m_w = 125g = 0.125kg$

$T_1 = 0°C$

$T_2 = ?$

$H = 75kJ$

$c_w = 4.2kJ/kg.°C$

$L_v = 2260kJ/kg$

First, we calculate the heat needed to convert that mass of water to steam;

H = mc_w(T2-T1)

H = 0.125×4.2×100 = 52.5kJ

Therefore, 22.5 (75-52.5) kJ will be left unused

Then we calculate if the heat left is sufficient to convert the water at 100°C to steam at 100°C

H = mL_v = 0.125 × 2260 = 282.5kJ

The heat left (22.5kJ) is not sufficient to change the state of the mass of water from liquid to steam

Therefore the final temperature of that mass of water is 100°C.