Question

A 254-rad dose of radiation is administered to a patient in an effort to combat a cancerous growth. Assume that all of the energy deposited is absorbed by the growth. (Use

4186 J/kg · °C

for the specific heat of water.)

(a) Calculate the amount of energy delivered per unit mass.
J/kg

(b) Assuming the growth has a mass of 0.25 kg and a specific heat equal to that of water, calculate its temperature rise.

Your response differs from the correct answer by more than 100%.°C

Answer 1

a)

rad = A unit of absorbed dose of ionizing radiation, corresponding to the absorption of 0.01 joule per kilogram of absorbing material

254-rad = 254 x 0.01 J/kg = 2.54 J/kg

b)

Mass of tumour = 0.25 kg.

absorbed dose = 0.25 x 2.54 J = 0.635 J

The specific heat of water = 4.18J/gC

1 J will raise the temperature of 1 g of water by 1/4.18 C

0.635 J will raise the temperature of 1g of tumour by 0.635/4.18 deg C

0.635 J will raise the temperature of 250 g of tumour by 0.635/(250*4.18) deg C

= 0.000608 degrees C