A lump of ice at -10 degree celsius weighing 0.80 kg is dropped into water at 0 degree celsius. If 0.005 kg of water freeze. Calculate the specific heat of ice.
I may have missed something but I don't get the right answer of 2.08 J/g*C. Please check your numbers.
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To calculate the specific heat of ice, we can use the formula:
Q = m * c * ΔT
where:
Q = heat transferred (in Joules)
m = mass of the substance (in kg)
c = specific heat capacity (in J/kg·C°)
ΔT = change in temperature (in C°)
First, let's calculate the heat transferred, Q:
Q = m * c * ΔT
Since 0.005 kg of water freezes, we know the heat transferred is equal to the heat lost by the ice. So we can write:
Q = m_water * c_water * ΔT_water
where:
m_water = mass of water that freezes (in kg)
c_water = specific heat capacity of water (in J/kg·C°)
ΔT_water = change in temperature of water (in C°)
In this case, calculate:
Q = 0.005 kg * c_water * 0
Note that the change in temperature of the water is 0°C because it reaches the freezing point of 0°C.
On the other hand, we know that the heat transferred from the ice to the water is the amount of heat required to change the ice from -10°C to 0°C. So we can write:
Q = m_ice * c_ice * ΔT_ice
where:
m_ice = mass of ice (in kg)
c_ice = specific heat capacity of ice (in J/kg·C°)
ΔT_ice = change in temperature of ice (in C°)
In this case, calculate:
Q = 0.80 kg * c_ice * (0°C - (-10°C))
Now, we have two equations for Q. Since the heat transferred is equal in both cases, we can equate the two expressions for Q:
0.005 kg * c_water * 0 = 0.80 kg * c_ice * (0 - (-10))
Simplifying:
0 = 8 * c_ice
Now solve for c_ice:
c_ice = 0 / 8
c_ice = 0
Therefore, the specific heat of ice is 0 J/kg·C°.
To calculate the specific heat of ice, we need to use the concept of heat transfer. The heat lost by the ice will be equal to the heat gained by the water that freeze. We can calculate this using the formula:
Q = mcΔT
Where:
Q is the heat transferred (in Joules)
m is the mass (in kilograms)
c is the specific heat capacity (in J/kg°C)
ΔT is the change in temperature (in °C)
Given data:
Initial temperature of the ice, Ti = -10°C
Final temperature of the ice, Tf = 0°C
Mass of the ice, m_ice = 0.80 kg
Mass of the water that freezes, m_water = 0.005 kg
First, let's calculate the heat lost by the ice:
Q_ice = m_ice * c_ice * (Tf - Ti)
Since the water is freezing, the heat gained by the water is equal to the heat lost by the ice:
Q_ice = Q_water
Therefore, we can rewrite the equation as:
m_ice * c_ice * (Tf - Ti) = m_water * L
Where L is the latent heat of fusion for water (amount of heat required to freeze 1 kg of water at 0°C, which is 333,500 Joules/kg).
Rearranging the equation, we get:
c_ice = (m_water * L) / (m_ice * (Tf - Ti))
Now, let's substitute the given values and calculate:
c_ice = (0.005 kg * 333,500 J/kg) / (0.80 kg * (0°C - -10°C))
c_ice = (0.005 * 333,500) / (0.80 * 10)
c_ice = 4,168.75 / 8
c_ice ≈ 520.75 J/kg°C
Therefore, the specific heat of ice is approximately 520.75 J/kg°C.