A beaker contains 50.0 mL of water at 90.0°C. How many calories must be added to raise the temperature of the water to its boiling point (100°C)?
A) 50.0 cal
B) 1.00*10^3
C) 100.0 cal
D) 5.00*10^2
I think the answer is B, but i also think im wrong...
Now we have a heat capacity question
mass of 50 mL = 50 grams
change in temp = 10 degrees C
specific heat capacity of water = 1 cal/gram deg C
so
heat in = 1 cal/g C * 50 g * 10 C
= 500 Calories
= 5 * 10^2 Calories
oh... thnx again Damon! I'll write this 1 down just in case i need it again.
To calculate the calories needed to raise the temperature of the water from 90.0°C to 100°C, we need to use the specific heat capacity of water. The specific heat capacity of water is approximately 1 calorie per gram per degree Celsius.
First, we need to calculate the mass of water in the beaker. The density of water is approximately 1 gram per milliliter. So, the mass of water is 50.0 grams.
Next, we need to calculate the temperature change. The temperature change is the final temperature (100°C) minus the initial temperature (90.0°C). The temperature change is 10.0°C.
Now we can calculate the amount of heat required using the formula:
Heat (calories) = Mass (grams) × Temperature Change (°C) × Specific Heat Capacity (cal/g °C)
Heat = 50.0 g × 10.0 °C × 1 cal/g °C = 500 calories.
Therefore, the correct answer is D) 5.00*10^2.
To find the number of calories required to raise the temperature of the water from 90.0°C to 100°C, you can use the equation:
Q = mcΔT
Where:
Q = heat energy (in calories)
m = mass of the water (in grams)
c = specific heat capacity of water (1 cal/g°C)
ΔT = change in temperature (in °C)
In this case, you need to determine the value of Q. Since you have the volume of water given (50.0 mL), you can use the density of water to find the mass.
The density of water is approximately 1 g/mL. Therefore, the mass can be calculated as follows:
mass = volume × density = 50.0 mL × 1 g/mL = 50.0 g
Now that we have the mass, we can substitute the values into the equation:
Q = mcΔT = 50.0 g × 1 cal/g°C × (100°C - 90.0°C) = 500 cal
The answer is therefore D) 5.00 * 10^2, which corresponds to 500 calories.