Question

200g of water is heated to a temperature

of 90C from a starting temperature of 15C. The specific heat of water is 4.181 J/g°C. How much heat energy was added to the water? Please give the answer without units. Equation: q = m x c x ΔT

A) 6300
B) 4158
C) 12540
D) 5238

Answer 1

To find the heat energy added to the water, the formula q = m x c x ΔT was used. The mass of water is 200 g, specific heat is 4.181 J/g°C, and the temperature change is 75°C. The heat energy added to the water is 62715 J, which is not one of the provided answer choices.

Step-by-step explanation:

To calculate the amount of heat energy that was added to the water, we use the formula q = m x c x ΔT, where q is the heat energy in joules, m is the mass of the water in grams, c is the specific heat capacity in J/g°C, and ΔT is the change in temperature.

Here, m = 200 g, c = 4.181 J/g°C, and ΔT is the change in temperature from 15°C to 90°C, which is 90°C - 15°C = 75°C. So, we calculate q as follows:

q = 200 g × 4.181 J/g°C × 75°C = 62715 J

So, the correct answer is none of the provided options. The amount of heat energy added is 62715 J, which is not listed in the provided choices.