How many grams of water when supplied with 348 J of heat will gain a temperature of 5.2oC?
Well, water you waiting for? Let's dive right in and calculate this!
To find out how many grams of water will gain a temperature of 5.2oC when supplied with 348 J of heat, we need to use the formula:
q = m * c * ΔT
Where:
- q represents the heat energy in Joules
- m represents the mass of water in grams
- c represents the specific heat capacity of water in J/g°C
- ΔT represents the change in temperature in °C
First, let's rearrange the formula to solve for mass (m):
m = q / (c * ΔT)
Now, the specific heat capacity of water is approximately 4.18 J/g°C.
So, plugging in the numbers:
m = 348 J / (4.18 J/g°C * 5.2°C)
After crunching the numbers, you will find that the mass of water is approximately 16.5 grams.
So, around 16.5 grams of water will have a temperature increase of 5.2oC when given 348 J of heat.
To calculate the grams of water, we can use the formula:
q = m * c * ΔT
where:
q is the amount of heat (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g·°C)
ΔT is the change in temperature (in °C)
In this case, we are given:
q = 348 J
ΔT = 5.2°C
The specific heat capacity of water is approximately 4.18 J/g·°C.
Let's plug in the values and solve for m:
348 J = m * 4.18 J/g·°C * 5.2°C
Dividing both sides of the equation by (4.18 J/g·°C * 5.2°C):
348 J / (4.18 J/g·°C * 5.2°C) = m
m ≈ 16.87 grams
Therefore, approximately 16.87 grams of water will gain a temperature of 5.2°C when supplied with 348 J of heat.
To find the number of grams of water that will gain a temperature of 5.2°C when supplied with 348 J of heat, we need to use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy supplied in joules,
m is the mass of the substance in grams,
c is the specific heat capacity of the substance (for water it is approximately 4.18 J/g°C),
ΔT is the change in temperature in degrees Celsius.
In this case, we are given Q (348 J) and ΔT (5.2°C), and we need to solve for m.
Rearranging the formula, we have:
m = Q / (c * ΔT)
Substituting the given values, we get:
m = 348 J / (4.18 J/g°C * 5.2°C)
Calculating this expression gives us the mass of water.
m = 348 J / (21.736 g°C)
m ≈ 16.01 grams of water
Therefore, when supplied with 348 J of heat, approximately 16.01 grams of water will gain a temperature of 5.2°C.
q = mcdT.
348 = m*4.184 J/oC*5.2
Solve for m in grams.