You have two cups, one containing 100 g of ice at 0oC, the other containing 100 g of water at 80oC. You pour the hot water into the cup containing the ice. What do you wind up with?
A) 100 g of 0oC water and 100 g of 80oC water.
B) 50 g of 0oC ice and 50g of 0oC water.
C) 200 g of 0oC water.
D) 100 g of 0oC water.
I think the answer is D am I correct
Seems to me you ultimately need to find the total for both the mass of the H2O and the mean for temperature. You will initially find the temperature of the combination water decreasing and volume of the ice decreasing. Once out of the solid state, the liquid ice will increase in temperature the entire combination will approach the mean.
Before reaching the mean of the temperature, you will find that the temperature close to the ice will be lower than that in the liquid state water further from the ice. Close to the ice, you should expect to find 0 degrees C. What will be interesting for you to explore is how long at it will take for the transition to that mean to take place.
You have two identical substances that have similar properties except that one does not generally exist in the solid form above its freezing point. Each gram of pure 80 degree water (each ml) will need to dissipate 1 calorie of energy per degree to reach the 40 degree mean. Each gram of pure water ice will need to absorb 1 calorie per degree to for it to reach 40 degrees. Since it is not an instantaneous heat exchange, both the different parts of the combination will have different physical and thermal properties until they reach that mean.
To solve this question, we need to understand the concept of heat transfer and the specific heat capacity of water and ice.
When the hot water at 80oC is poured into the cup with ice at 0oC, heat is transferred from the hot water to the ice. This heat transfer causes the ice to melt and the temperature of the water to decrease.
To find the final state after heat transfer, we can use the equation:
Q = m*c*ΔT
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
For the ice, we can assume that the specific heat capacity is 2.09 J/g°C, and for water, it is approximately 4.18 J/g°C.
The hot water loses heat and cools down while the ice melts and heats up. The heat transferred will be equal for both substances, as it is a closed system.
Let's calculate the heat transferred from the hot water to the ice:
Q_ice = m_ice * c_ice * ΔT_ice
= 100 g * 2.09 J/g°C * (0°C - (-80°C))
= 100 g * 2.09 J/g°C * 80°C
= 16,720 J
To melt the ice, the heat transferred must be sufficient to meet the heat of fusion (also known as the latent heat of fusion), which is the energy required to change a substance from solid to liquid without changing its temperature. For water, the heat of fusion is 334 J/g.
Let's calculate the heat required to melt the ice:
Q_melt = m_ice * ΔH_fusion
= 100 g * 334 J/g
= 33,400 J
Since the heat transferred (16,720 J) is less than the heat required to melt the ice (33,400 J), the ice will not entirely melt. Instead, some of the ice will melt, and the remaining ice will cool down.
Therefore, the correct answer is option B) 50 g of 0oC ice and 50 g of 0oC water.