sally had 20g of water and it started at a temperature of 26 degrees celsius and sally cooled it to 22 degrees celsius. a)how much did the temperature of water change? (show work) B) how many calories did sally's water lose?
To determine the change in temperature, subtract the initial temperature from the final temperature:
Change in temperature = Final temperature - Initial temperature
In this case:
Change in temperature = 22 degrees Celsius - 26 degrees Celsius = -4 degrees Celsius
The temperature decreased by 4 degrees Celsius.
To calculate the number of calories lost by the water, you'll need to know the specific heat capacity of water, which is 1 calorie/gram•degree Celsius (cal/g•°C).
To find the calories lost, you'll need to use the equation:
Calories lost = Change in temperature (in degrees Celsius) * Mass of water (in grams) * Specific heat capacity of water (cal/g•°C)
Let's break down the calculation for Sally's water:
a) Change in temperature = -4 degrees Celsius (from earlier calculation)
b) Mass of water = 20 grams (given in the question)
c) Specific heat capacity of water = 1 cal/g•°C (known constant)
Now, substitute these values into the formula:
Calories lost = -4 degrees Celsius * 20 grams * 1 cal/g•°C
Calories lost = -80 calories
Therefore, Sally's water lost 80 calories. Note that the negative sign indicates a loss of calories.