Question

How much energy must be absorbed by water with a mass of 0.05 kg in order to raise the temperature from 30°C to 65°C? Note: Water has a specific heat of 4,190 J/kg °C.

Answer 1

Step-by-step explanation:

Use the calorimetry formula.

Q = mc∆T; where Q = heat energy, c = specific heat capacity, and ∆T = change in temperature

Known and Unknown

Q = ? J

m = 0.05 kg

c = 4190 J/kg•°C

T(i) = initial temperature = 30 °C

T(f) = final temperature = 65 °C

∆T = T(f) - T(i) = 65 °C - 30 °C = 35 °C

Calculate the heat energy.

Q = 0.05 kg × 4190 J/kg•°C × 35 °C = 7332.5 J