Question

The specific heat of liquid water is 4.18 j/g-°c. how many joules of heat are needed to raise the temperature of 7.25 g of water from 20.0 °c to 65.3 °C?

Answer 1

Approximately 1373.94 joules of heat energy are required to raise the temperature of 7.25 grams of water from 20.0 °C to 65.3 °C, calculated using the specific heat capacity of water and the formula for heat energy transfer.

Step-by-step explanation:

The student is asking how many joules of heat energy are required to raise the temperature of 7.25 grams of water from 20.0 °C to 65.3 °C. Using the formula Q = mcΔT, where Q is the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature, we can calculate the required heat energy.

First, we find the change in temperature: ΔT = (65.3 °C - 20.0 °C) = 45.3 °C. Next, we use the specific heat capacity of water, c = 4.184 J/g °C, and the mass of water, m = 7.25 g, to calculate Q.

So, Q = (7.25 g) × (4.184 J/g °C) × (45.3 °C) = 1373.9356 J. Therefore, 1373.94 joules of heat energy are needed to raise the temperature of 7.25 grams of water from 20.0 °C to 65.3 °C.