A system is provided `50` joule of heat and the change in internal energy during the process is `60 J`. Magnitude of work done on the system is:

Answer

To solve the problem, we will use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). The formula can be expressed as:

\[ \Delta U = Q - W \]

Where:
- ΔU = Change in internal energy
- Q = Heat added to the system
- W = Work done by the system

In this case, we need to find the magnitude of work done on the system, which can be rearranged from the equation above:

\[ W = Q - \Delta U \]

### Step-by-step Solution:

1. **Identify the given values**:
- Heat provided to the system (Q) = 50 Joules
- Change in internal energy (ΔU) = 60 Joules

2. **Substitute the values into the equation**:
\[ W = Q - \Delta U \]
\[ W = 50 \, \text{J} - 60 \, \text{J} \]

3. **Calculate the work done**:
\[ W = 50 \, \text{J} - 60 \, \text{J} = -10 \, \text{J} \]

4. **Interpret the result**:
The negative sign indicates that work is done on the system. However, since the question asks for the magnitude of work done, we take the absolute value:
\[ \text{Magnitude of work done} = |W| = 10 \, \text{J} \]

### Final Answer:
The magnitude of work done on the system is **10 Joules**.