If 10 calories of heat are added to two grams of liquid water what is the temperature of the water
To determine the change in temperature of the water when 10 calories of heat are added, you need to use the specific heat capacity formula:
Q = mcΔT
Where:
Q = amount of heat added (in calories)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in cal/g°C)
ΔT = change in temperature (in °C)
In this case, we have:
Q = 10 calories (heat added)
m = 2 grams (mass of water)
To find the specific heat capacity of water, the value is approximately 1 calorie/gram°C.
Plugging these values into the formula:
10 calories = 2 grams * 1 calorie/gram°C * ΔT
Simplifying the equation:
10 calories = 2 calories/°C * ΔT
Divide both sides of the equation by 2 calories/°C:
10 calories / (2 calories/°C) = ΔT
This gives us:
5°C = ΔT
So, when 10 calories of heat are added to two grams of liquid water, the temperature of the water will increase by 5 degrees Celsius.
To calculate the temperature change of the water, you can use the formula:
Q = m * c * ΔT
Where:
Q = Heat energy (calories)
m = Mass of the water (grams)
c = Specific heat capacity of water (calories/gram°C)
ΔT = Change in temperature (°C)
In this case, you have 10 calories of heat energy added to 2 grams of water. The specific heat capacity of water is approximately 1 calorie/gram°C.
Substituting the given values into the formula:
10 calories = 2 grams * 1 calorie/gram°C * ΔT
Simplifying the equation:
10 calories = 2 calories/°C * ΔT
Dividing both sides of the equation by 2 calories/°C:
ΔT = 10 calories / 2 calories/°C
Calculating the result:
ΔT = 5 °C
Therefore, the temperature of the water will increase by 5 degrees Celsius.
If you don't know the temperature at the start you can't know the final T; however, you can calculate delta T.
q = m*c*delta T
10 = 2 x 1 x dT. Solve for delta T. I get T will be 5 degrees C warmer than where it started.