Suppose that 90 g of hot water at a temperature of 75 degrees C is poured into a cavity in a very large block of ice at 0 degrees C.
(a) What will be the final temperature of the water in the cavity?
(b) What mass of ice must melt in order for the water to cool down to the temperature found in Part (a)?
(a) If the ice block is really "very large", not all of it melts, and the final temperature remains 0 C. You end up with a large pool of 0 C liquid water including melted ice) in the ice block.
(b) If mass M of original ice melts,
Heat lost by poured water = Heat gained by melted ice
90g*75C*1.00cal/g*C = M (g)*80 (cal/g)
Solve for M
M = 84.4 g
To calculate the final temperature of the water in the cavity, we can use the principle of heat transfer. The heat lost by the hot water will be equal to the heat gained by the ice.
(a) Let's use the formula for heat transfer:
Q = m * c * ΔT
Where:
Q = heat transferred
m = mass
c = specific heat capacity
ΔT = change in temperature
First, let's calculate the heat lost by the hot water:
Q_lost = m_hot * c_hot * ΔT_hot
Given:
m_hot = 90 g
c_hot = 4.18 J/g°C (specific heat capacity of water)
ΔT_hot = 75 - 0 = 75°C
Q_lost = 90 g * 4.18 J/g°C * 75°C
Now, let's calculate the heat gained by the ice:
Q_gained = m_ice * c_ice * ΔT_ice
Given:
c_ice = 2.09 J/g°C (specific heat capacity of ice)
ΔT_ice = 0 - T_final (T_final is the final temperature of the water in the cavity)
We can assume that there is no phase change involved, so the specific heat capacity of water can be used for ice as well.
Q_gained = m_ice * c_ice * (-T_final)
Now, equating the two equations:
Q_lost = Q_gained
90 g * 4.18 J/g°C * 75°C = m_ice * 2.09 J/g°C * (-T_final)
Now, let's simplify the equation:
m_ice = (90 g * 4.18 J/g°C * 75°C) / (2.09 J/g°C * (-T_final))
Now, let's solve for the final temperature (T_final):
(b) We know that when ice melts into water at 0°C, no change in temperature occurs. So, the final temperature (T_final) will be 0°C.
If T_final = 0°C, the equation becomes:
m_ice = (90 g * 4.18 J/g°C * 75°C) / (2.09 J/g°C * (0°C))
Now, let's calculate the value of m_ice:
m_ice = (90 g * 4.18 J/g°C * 75°C) / (2.09 J/g°C * (0°C))
m_ice = (90 g * 4.18 J) / (2.09 J)
m_ice = 180 g
Therefore, to cool the water from 75°C to 0°C, 180 g of ice must melt.
To find the final temperature of the water in the cavity (a) and the mass of ice that must melt (b), we can apply the principles of heat transfer and the specific heat capacities of water and ice.
(a) To find the final temperature of the water in the cavity, we can use the principle of heat transfer:
The heat lost by the water = The heat gained by the ice.
The heat gained by the ice can be calculated using the specific heat capacity of ice (cice = 2.09 J/g°C), the mass of ice melted (mice), and the change in temperature from 0°C to the final temperature (Tf):
Heat gained by ice = mice * cice * Tf
The heat lost by the water can be calculated using the specific heat capacity of water (cwater = 4.18 J/g°C), the mass of the water (mwater = 90 g), and the change in temperature from 75°C to the final temperature (Tf):
Heat lost by water = mwater * cwater * (75°C - Tf)
Since the heat lost by the water is equal to the heat gained by the ice, we can set these two equations equal:
mwater * cwater * (75°C - Tf) = mice * cice * Tf
Solving this equation will give us the final temperature (Tf) of the water in the cavity.
(b) To find the mass of ice that must melt for the water to cool down to the temperature found in part (a), we can use the specific latent heat of fusion of ice (Lf = 334 J/g).
The heat required to melt the ice is given by:
Heat required = mice * Lf
Since the total heat lost by the water is equal to the heat required to melt the ice, we can set these two equations equal:
mwater * cwater * (75°C - Tf) = mice * Lf
Solving this equation will give us the mass of ice (mice) that must melt.
By solving these equations, we can find the final temperature of the water in the cavity (a) and the mass of ice that must melt (b).