Answer:
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
