In the problem about heating a 200 g copper pipe with a Bunsen burner supplying 7410 J (with copper specific heat capacity 390 J/kg °C), what is the change in temperature of the pipe assuming all the energy heats the pipe?

What you are solving for

You are told how much thermal energy $Q$ the Bunsen burner supplies, and you want the temperature change $\Delta T$ of the copper pipe, assuming no energy is lost to the surroundings.

Convert the mass into kilograms

Specific heat capacity is given in $\text{J}/(\text{kg}\,^{\circ}\text{C})$, so mass must be in kg:

$$200\text{ g} = 0.200\text{ kg}$$

Rearrange the specific heat equation

Start with:

$$Q = mc\Delta T$$

Solve for $\Delta T$:

$$\Delta T = \frac{Q}{mc}$$

Substitute values and calculate

Use $Q = 7410\text{ J}$, $m = 0.200\text{ kg}$, $c = 390\,\text{J}/(\text{kg}\,^{\circ}\text{C})$:

$$\Delta T = \frac{7410}{0.200 \times 390} = \frac{7410}{78} = 95\,^{\circ}\text{C}$$

Quick reasonableness check

A larger mass or a larger specific heat capacity would make $mc$ bigger and $\Delta T$ smaller. Here, $mc = 78\,\text{J}/^{\circ}\text{C}$, so each $78\text{ J}$ raises the pipe by about $1^{\circ}\text{C}$, and $7410\text{ J}$ gives about $95^{\circ}\text{C}$, which is consistent.