A small pebble is heated and placed in a foam cup calorimeter containing water at 25 °C. The water reaches a maximum temperature of 27 °C. If the pebble released 532.1 J of heat to the water, what mass (in g) of water was in the calorimeter?
heat lost by pebble + heat gained by water = 0
heat lost by pebble = 532.1 J
heat gained by water = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Set q = heat gaine by water and solve for mass H2O.
To solve this problem, we can use the formula for heat transfer:
Q = (m × c × ΔT)
Where:
Q = heat transfer (in J)
m = mass of the substance (in g)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
Given values:
Q = 532.1 J
ΔT = 27 °C - 25 °C = 2 °C
Assuming the specific heat capacity of water is approximately 4.18 J/g°C, we can rearrange the formula to solve for the mass (m) of water:
m = Q / (c × ΔT)
Substituting the given values into the formula:
m = 532.1 J / (4.18 J/g°C × 2 °C)
m ≈ 63.96 g
Therefore, the mass of water in the calorimeter is approximately 63.96 g.
To find the mass of water in the calorimeter, you need to use the equation for heat transfer:
q = mcΔT
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
- ΔT is the change in temperature
In this case, the pebble released 532.1 J of heat to the water, and the water temperature increased from 25 °C to 27 °C.
The specific heat capacity of water is approximately 4.18 J/g °C.
Substituting the given values into the equation:
532.1 J = m * 4.18 J/g °C * (27 °C - 25 °C)
Now, we can solve for the mass (m):
532.1 J = m * 4.18 J/g °C * 2 °C
Rearranging the equation:
m = 532.1 J / (4.18 J/g °C * 2 °C)
Calculating:
m = 532.1 J / 8.36 J/g
m ≈ 63.82 g
Therefore, the mass of water in the calorimeter is approximately 63.82 grams.