A 500 mL bottle of spring water, which is at room temp. of25oC, is poured over 120g of ice that is at-8oC. What will be the final temperature of the waterwhen all the ice has melted, assuming that it is in an insulatedcontainer that does not change temperature?

T = 3.95 ° C

Explanation:

Given:

- The volume of water bottle V = 500 mL

- The initial temperature of water bottle T_i,w = 25° C

- The mass of ice m_ice = 120 g

- The initial temperature of ice T_i,ice = -8° C

- The specific capacity of ice C_ice = 2.11 J/gC

- The specific capacity of water C_w = 4.184 J/gC

- The heat of fusion for water L_f = 333.55 J/g

- Density of water p_w = 1.0 g/cm^3

Find:

What will be the final temperature of the water when all the ice has melted

Solution:

- Assuming no heat loss from the system. All the heat from the bottle of water is transferred to the block of ice.

- The Heat required for block of ice to reach its melting point T_m = 0° C.

            Q_1 = m_ice*C_ice*( T_m - T_i,ice )

            Q_1 = 120*2.11*( 0 - (-8) )

            Q_1 = 2025.6 J

- The Heat required for block of ice to melt:

            Q_2 = m_ice*L_f

            Q_2 = 120*333.55

            Q_2 = 40026 J

- The Heat required for block of ice-water to reach Equilibrium T.

            Q_3 = m_ice*C_w*( T - 0 )

            Q_3 = 120*4.184*T

            Q_3 = 502.08*T

- Now for water, the amount of heat lost would be:

            Q_4 = p_w*V*C_w*( T - 25 )

            Q_4 = 1.0 g/cm^3*500*4.184* ( T - 25 )

            Q_4 = 2092*( T - 25 )

- Setting up an energy balance:

            Q_4 = Q_1 + Q_2 + Q_3

            2092*( 25-T ) = 2025.6 + 40026 + 502.08*T

            1589.92*T = 2025.6 + 40026 - 52300

            2594.08*T = 10248.4

             T = 3.95 ° C

d. platinum would have the greatest temperature increase because it has the lowest specific heat.

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