A 44 g piece of ice at 0 degrees celsius is added to a sample of water at 7 degrees celsius. all of the ice melts and the temperature of the water decreases to 0 degrees celsius. How many grams of water were in the sample?
44g x 80.0 cal/g )+( 7 x 0.50 cal/g x 1.00 cal/g) =3520 cal +3.5 cal/g = 3523.5 cal
To find out how many grams of water were in the sample, we can use the principle of heat transfer. The heat lost by the water as it cools down to 0 degrees Celsius is gained by the ice as it melts.
First, we need to calculate the amount of heat lost by the water. We can use the formula:
Q = m * c * ΔT
Where:
Q = heat energy in Joules
m = mass of water in grams
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (7°C - 0°C = 7°C)
Q = m * 4.18 * 7
Now, we know that the heat gained by the ice is equal to the heat lost by the water. Since the ice is at 0 degrees Celsius, it will absorb heat while melting, but its temperature will remain constant.
The heat absorbed by the ice can be calculated using the formula:
Q = mL
Where:
Q = heat energy in Joules
m = mass of ice in grams
L = latent heat of fusion of ice (334 J/g)
Q = 44 * 334
Since Q is the same for both the water and the ice, we can set the two equations equal to each other:
m_water * 4.18 * 7 = 44 * 334
Now, we can solve for m_water to find the mass of water in grams.