How much ice at zero degrees Celcius (in grams) would have to melt to lower the temperature of 351 of water from 25 to 5? (Assume the density of water is 1.0 .)
can not do this without units
x = grams of ice
x times heat of fusion of water in cal/gram = grams of water * specific heat of water in cal/deg gm * (25-5)
x = grams of ice
x times heat of fusion of water in cal/gram = grams of water * specific heat of water in cal/deg gm * (25-5)
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That assumes that the ice is separated from the water. If they mix then the ice not only has to melt but also be raised from 0 to 5. Then:
x = grams of ice
x times heat of fusion of water in cal/gram + x *specific heat of water * (5-0) = grams of water * specific heat of water in cal/deg gm * (25-5)
To find out how much ice would have to melt, we can use the equation:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the substance (in this case, ice)
c is the specific heat capacity of the substance (in this case, ice is 2.09 J/g°C)
ΔT is the change in temperature
First, we need to calculate the heat Q transferred from the water, which can be calculated using the equation:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the substance (in this case, water)
c is the specific heat capacity of the substance (in this case, water is 4.18 J/g°C)
ΔT is the change in temperature
Substituting the given values:
Q = 351 g * 4.18 J/g°C * (5°C - 25°C)
Simplifying the equation:
Q = 351 g * 4.18 J/g°C * (-20°C)
Q = -29508 J
The negative sign indicates that heat is being transferred out of the water.
Now, we can calculate the mass of ice that needs to melt to transfer this amount of heat:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the substance (in this case, ice)
c is the specific heat capacity of the substance (in this case, ice is 2.09 J/g°C)
ΔT is the change in temperature
Substituting the values:
-29508 J = m * 2.09 J/g°C * (0°C - 0°C)
Simplifying the equation:
-29508 J = m * 2.09 J/g°C * 0°C
Since the temperature change is 0°C, the equation becomes:
-29508 J = m * 0 J/g
We can see that the mass of ice that needs to melt is zero grams since any mass multiplied by zero is zero. Therefore, no ice needs to melt to lower the temperature of the water from 25°C to 5°C.