Question

An ice cube at 0.00 ∘C with a mass of 25.0 g is placed into 650.0 g of water, initially at 29.0 ∘C, in an insulated container. Assuming that no heat is lost to the surroundings, what is the temperature of the entire water sample after all of the ice has melted?

Answer 1

Heat lost by warm water = heat neede to melt ice + heat needed to warm water whic was once ice

$$m_w\times c_w\times(T_w-T_f) = m_{ice}\times L_f+m_{ice}\times c_w\times(T_f-0)$$

where

$$c_w = 4.184 \frac{J}{g^\circ C}$$

$$L_f = 334 \frac{J}{g}$$

$$650.0\times 4.184\times (29.0-T_f) = 25.0\times 334 + 25.0 \times 4.184 (T_f - 0)$$

$$78868.4 -2719.6\times T_f = 8350 + 104.6\times T_f$$

$$2824.2 \times T_f = 70518.4$$

$$T_f = 25.0$$