a 52.0 gram sample or water increases in temperature due to the addition of a hot metal from 17.7 degrees C to 24.4 degrees C.
How many calories has the water absorbed?
how many calories did the metal lose?
To determine how many calories the water absorbed and how many calories the metal lost, we need to calculate the heat gained/absorbed by the water and the heat lost by the metal.
First, let's calculate the heat gained/absorbed by the water using the formula:
Q = mcΔT
Where:
Q = heat gained/absorbed by the water
m = mass of the water (in grams)
c = specific heat capacity of water (approximately 1 calorie/gram °C)
ΔT = change in temperature of the water (in °C)
Given:
m (mass of water) = 52.0 grams
ΔT (change in temperature) = 24.4 °C - 17.7 °C = 6.7 °C
Substituting the given values into the formula:
Q = (52.0 g)(1 cal/g °C)(6.7 °C)
Calculating this gives us:
Q = 349.6 calories
Therefore, the water absorbed 349.6 calories.
To determine the calories lost by the metal, we assume that the heat lost by the metal is equal to the heat gained by the water (according to the Law of Conservation of Energy):
Q (water) = Q (metal)
Therefore, the metal lost 349.6 calories.
So, the water absorbed 349.6 calories, and the metal lost 349.6 calories.