An X-ray tube is operating at 150 kV and 10 mA. If only 1% of the electric power supplied is converted into X-rays, the rate at which the target is heated in calories per second is

Options

A. 3.55
B. 35.5
C. 355
D. 3550

Explanation

To solve the problem, we need to calculate the rate at which the target in an X-ray tube is heated in calories per second. We know that the X-ray tube operates at a voltage of 150 kV and a current of 10 mA, and that only 1% of the electric power supplied is converted into X-rays.

### Step-by-Step Solution:

1. **Convert Voltage and Current to Standard Units:**
- Voltage (V) = 150 kV = \(150 \times 10^3\) volts
- Current (I) = 10 mA = \(10 \times 10^{-3}\) amperes

2. **Calculate the Total Power Supplied:**
- Power (P) is given by the formula:
\[ P = V \times I \]
- Substituting the values:
\[ P = (150 \times 10^3) \times (10 \times 10^{-3}) = 1500 \text{ watts} \]

3. **Determine the Power Converted to X-rays:**
- Since only 1% of the power is converted into X-rays, the power converted to X-rays is:
\[ P_{\text{X-rays}} = 0.01 \times P = 0.01 \times 1500 = 15 \text{ watts} \]

4. **Calculate the Power Used to Heat the Target:**
- The remaining power (99% of the total power) is used to heat the target:
\[ P_{\text{heating}} = 0.99 \times P = 0.99 \times 1500 = 1485 \text{ watts} \]

5. **Convert Watts to Calories per Second:**
- To convert the heating power from watts to calories per second, we use the conversion factor \(1 \text{ watt} = \frac{1 \text{ joule}}{1 \text{ second}} = \frac{1}{4.18} \text{ calories}\):
\[ P_{\text{heating in calories}} = \frac{1485}{4.18} \approx 355 \text{ calories per second} \]

### Final Answer:
The rate at which the target is heated is approximately **355 calories per second**.