If meat is irradiated with 2000.0 Gy of \(\mathrm{x}\) -rays, most of the bacteria are killed and the shelf life of the meat is greatly increased. (a) How many 100.0 -keV photons must be absorbed by a \(0.30-\mathrm{kg}\) steak so that the absorbed dose is 2000.0 Gy? (b) Assuming steak has the same specific heat as water, what temperature increase is caused by a 2000.0 -Gy absorbed dose?

First, let's calculate the total energy absorbed by the steak for the 2000 Gy absorbed dose. The relationship between absorbed dose (D), energy (E), and mass (m) is given by: \(D = \frac{E}{m}\) We are given: Absorbed dose (D) = 2000 Gy Mass of steak (m) = 0.30 kg Rearranging the equation for energy (E), we get: \(E = D \times m\) Now plug in the values: \(E = 2000 \ \text{Gy} \times 0.30 \ \text{kg} = 600 \ \text{J}\) So, the steak must absorb 600 J of energy.

Now, let's find the number of photons required to deliver this amount of energy. We know that each photon has energy 100 keV. First, convert this energy to joules: \(100 \ \text{keV} = 100 \times 10^3 \ \text{eV} = 100 \times 10^3 \times 1.6 \times 10^{-19} \ \text{J} = 1.6 \times 10^{-14} \ \text{J}\) Next, we can find the number of photons (n) needed using the equation: \(n = \frac{E_\text{total}}{E_\text{photon}}\) Where: \(E_\text{total} = 600 \ \text{J}\) (total energy needed) \(E_\text{photon} = 1.6 \times 10^{-14} \ \text{J}\) Plugging in the values: \(n = \frac{600}{1.6 \times 10^{-14}} \approx 3.75 \times 10^{16}\) So, approximately \(3.75 \times 10^{16}\) 100 keV photons must be absorbed by the steak.

Finally, let's calculate the temperature increase caused by the 2000 Gy absorbed dose, assuming the steak has the same specific heat as water. The specific heat of water (c) is \(4.18 \ \text{J} / \text{g} \cdot \text{°C}\). Using the heat formula: \(\Delta Q = m \times c \times \Delta T\) We are given: \(\Delta Q = 600 \ \text{J}\) (energy absorbed) Mass (m) = \(0.30 \ \text{kg} \times 1000 = 300 \ \text{g}\) We need to find the temperature change (\(\Delta T\)): \(\Delta T = \frac{\Delta Q}{m \times c}\) Plugging in the values: \(\Delta T = \frac{600 \ \text{J}}{300 \ \text{g} \times 4.18 \ \text{J/g} \cdot \text{°C}} \approx 0.478 \ \text{°C}\) So, the temperature increase of the steak due to a 2000 Gy absorbed dose is approximately 0.478 °C.