1. A water of mass 7kg at a temperature of 65°c is added to 14kg of water at 10°c. calculate the final temperature of the mixture. [SHC of water Cw=4,200J/kg.K]
2. How much quantity of heat energy is required to change 7.4kg of ice at 0°C into water at the same temperature? [Lf=306,000J/kg]
3. How much quantity of heat energy is required to change 8.0kg of ice at 0°C into water at the same temperature? [LV=2,430,000J/kg]
1. First, calculate the total heat lost by the hot water and gained by the cold water:
Q = m1 * Cw * (Tf - Th) + m2 * Cw * (Tf - Tc)
Q = 7 * 4200 * (Tf - 65) + 14 * 4200 * (Tf - 10)
Next, set the total heat lost equal to the total heat gained and solve for Tf:
7 * 4200 * (Tf - 65) + 14 * 4200 * (Tf - 10) = 0
29400Tf - 294000 + 58800Tf - 588000 = 0
88200Tf = 882000
Tf = 10°C
Therefore, the final temperature of the mixture will be 10°C.
2. The heat energy required to change ice at 0°C into water at the same temperature can be calculated using the formula:
Q = m * Lf
Q = 7.4 * 306000
Q = 2,264,400 J
Therefore, 2,264,400 J of heat energy is required.
3. The heat energy required to change ice at 0°C into water at the same temperature can be calculated using the formula:
Q = m * Lv
Q = 8.0 * 2430000
Q = 19,440,000 J
Therefore, 19,440,000 J of heat energy is required.