A 51-g piece of ice at 0 C is added to a sample of water at 7-C all of the ice melts and the temp of the water decreases to 0 .If you used the 8200 Khan you expand in energy in one day to heat 47000g of water at 27-C what would be the rise in temp? What would the new temp of the water?
To calculate the rise in temperature and the new temperature of the water after using energy from an 8200 kJ heat source to heat 47000 g of water at 27 °C, we need to consider the specific heat capacity and the heat transfer equations.
First, let's determine the heat required to raise the temperature of the water. The heat transfer equation is given by:
Q = mcΔT
Where:
Q is the heat energy transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
The specific heat capacity of water is approximately 4.18 J/g°C.
1. Calculate the initial heat of the water:
Q1 = mcΔT
= (47000 g) * (4.18 J/g°C) * (27°C - 7°C)
= 47000 g * 4.18 J/g°C * 20°C
= 394,9600 J
Next, let's calculate the heat required to melt the ice and the subsequent temperature decrease of the water.
2. Calculate the heat required to melt the ice:
Q2 = mL
= (51 g) * (334 J/g)
= 17,034 J
Since the ice is at 0 °C and the water temperature decreases to 0 °C after the ice melts, the total heat energy required for this step is 17,034 J.
Now, let's calculate the total heat energy required:
Total heat energy = Q1 + Q2
= 394,960 J + 17,034 J
= 412,994 J
Finally, let's calculate the rise in temperature and the new temperature of the water:
Rise in temperature = (Total heat energy) / (mc)
= 412,994 J / (47000 g * 4.18 J/g°C)
≈ 2.20 °C
The rise in temperature would be approximately 2.20 °C.
New temperature of the water = Initial temperature + Rise in temperature
= 27 °C + 2.20 °C
= 29.20 °C
Therefore, the new temperature of the water would be approximately 29.20 °C after using the energy from the 8200 kJ heat source.