what mass of water at 15C can be cooled 1C by heat necessary to melt 185g of ice at 0C
mass H2O x specific heat water x (-1) + mass ice x heat fusion = 0.
To solve this problem, we need to calculate the amount of heat required to melt 185g of ice at 0°C, and then use that value to determine the mass of water that can be cooled by 1°C.
Step 1: Calculate the heat required to melt the ice:
The heat required to melt a substance can be calculated using the formula Q = m * L, where Q is the heat, m is the mass, and L is the latent heat of fusion.
The latent heat of fusion of ice is 334 Joules per gram (J/g).
Q = m * L
Q = 185g * 334 J/g
Q = 61,790 Joules (J)
Step 2: Calculate the mass of water that can be cooled:
The specific heat capacity of water is 4.18 J/g°C. To calculate the mass of water that can be cooled by 1°C using 61,790 Joules, we can use the formula:
Q = m * c * ΔT
Where Q is the heat, m is the mass, c is the specific heat capacity of water, and ΔT is the change in temperature.
Rearranging the formula, we get:
m = Q / (c * ΔT)
m = 61,790 J / (4.18 J/g°C * 1°C)
m = 14,804 g
Therefore, the mass of water that can be cooled by 1°C using the heat necessary to melt 185g of ice at 0°C is 14,804 grams.
To find the mass of water that can be cooled by the heat necessary to melt 185g of ice at 0°C, we need to calculate the heat exchanged in both processes.
First, let's consider the heat required to melt the ice. The heat necessary to change the phase of a substance can be calculated using the formula:
Q = m * L
where Q is the heat, m is the mass, and L is the latent heat of fusion for ice (which is 334 J/g).
So, for melting 185g of ice, the heat exchanged is:
Q_ice = m_ice * L_ice
Q_ice = 185g * 334 J/g
Q_ice = 61690 J
Next, let's calculate the heat exchanged when cooling water from 15°C to 14°C. The heat exchanged can be determined using the formula:
Q = m * C * ΔT
where Q is the heat, m is the mass, C is the specific heat capacity of water (which is approximately 4.18 J/g°C), and ΔT is the temperature change (which is 1°C in this case).
So, for cooling water with an unknown mass, the heat exchanged is:
Q_water = m_water * C_water * ΔT
Q_water = m_water * 4.18 J/g°C * 1°C
Q_water = 4.18 m_water J
Since the total heat exchanged must be equal to zero (since the heat lost by the water is gained by melting the ice):
Q_ice + Q_water = 0
61690 J + 4.18 m_water J = 0
Solving for m_water:
4.18 m_water J = -61690 J
m_water = -61690 J / 4.18 J/g
m_water ≈ -14794 g
Since mass cannot be negative, it means that the heat released by melting 185g of ice at 0°C is enough to cool approximately 14,794g of water from 15°C to 14°C.