What temperature change in Celsius degrees is produced when 800.00 calories are absorbed by 120.0 grams of water?
q = mcdT
800 = 120g x 1 cal/g x delta T
Solve for delta T.
To calculate the temperature change in Celsius degrees, we can use the specific heat capacity formula:
Q = mcΔT
Where:
Q = Heat energy absorbed/released (in calories)
m = Mass of the substance (in grams)
c = Specific heat capacity of the substance (in calories/gram°C)
ΔT = Temperature change (in °C)
In this case, we have:
Q = 800.00 calories
m = 120.0 grams
The specific heat capacity of water is approximately 1 calorie/gram°C. So, c = 1.
Now we can rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
Substituting the given values:
ΔT = 800.00 calories / (120.0 grams * 1 calorie/gram°C)
Calculating:
ΔT = 800.00 calories / 120.0 calories/°C
ΔT = 6.67 °C
Therefore, a temperature change of approximately 6.67 °C is produced when 800.00 calories are absorbed by 120.0 grams of water.
To find the temperature change in Celsius degrees, we can use the formula:
ΔT = Q / (m * c)
Where:
ΔT = change in temperature in Celsius degrees
Q = amount of heat absorbed or released in calories
m = mass of the substance in grams
c = specific heat capacity of the substance
In this case, we have:
Q = 800.00 calories
m = 120.0 grams
c = specific heat capacity of water, which is 1 calorie/gram/°C
Now we can substitute the given values into the formula:
ΔT = 800.00 calories / (120.0 grams * 1 calorie/gram/°C)
In this equation, the units of "calories" and "grams" cancel out, leaving us with:
ΔT = 800.00 / 120.0 °C
Simplifying the equation, we get:
ΔT = 6.67 °C
Therefore, when 800.00 calories are absorbed by 120.0 grams of water, the temperature change is 6.67 °C.