If 22 g of water is heated so that the water temperature rises from 18.1°C to 21.5°C, how many calories have been absorbed by the water?
The specific heat of water = 1.00 cal / (g · °C)
22g *(21.5-18.1)*1.00cal/g*C
When water is mixed with acetonitrile in a calorimeter, the temperature of the solution decreases because
To calculate the calories absorbed by the water, we can use the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat absorbed in calories
m is the mass of the water in grams
c is the specific heat of water in cal/(g·°C)
ΔT is the change in temperature in °C
Given:
m = 22 g (mass of water)
c = 1.00 cal/(g·°C) (specific heat of water)
ΔT = 21.5°C - 18.1°C = 3.4°C
Now, substitute the values into the formula:
Q = 22 g * 1.00 cal/(g·°C) * 3.4°C
Q = 74.8 calories
Therefore, 22 g of water absorbs 74.8 calories of heat.