Question

9,819 J were absorbed by a pond, sending its temperature rising from 15 C to 29 C. How much water was in the pond? (Specific heat of water=4.18J/gC)

Answer 1

In this question, we have a situation in which we have to use the Calorimetry formula, which is how much heat was released or absorbed (in Joules), after we had a change in temperature of a compound. The formula for this Calorimetry question is:

Q = mcΔT

Where:

Q = is energy as Heat, 9,819 J

m = mass in grams, this is what we want to find

c = is the specific heat capacity, 4.184

ΔT = the change in temperature, calculated as Final Temperature - Initial T, 29 - 15 = 14°C

Now we add these values into the formula:

9819 = m * 4.184 * 14

9819 = 58.576m

m = 9819/58.576

m = 167.6 grams of Water