Question

2) Calculate the temperature change for mercury if 0.160 kg of the metal absorb 1500 J of heat energy.

Mercury's specific heat is 0.14 J/g x°C.

Answer 1

`\displaystyle \Delta T & = 67\text{$^\circ$C}`

Step-by-step explanation:

Recall that the energy change of a substance is given by:

`\displaystyle q = mC\Delta T`

Where C is the substance's specific heat.

Substitute in known values and solve for ΔT:

`\displaystyle \begin{aligned} (1500\text{ J}) & = (0.160\text{ kg})\left(\frac{0.14\text{ J}}{\text{g-$^\circ$C}}\right)\left(\frac{1000 \text{ g}}{1\text{ kg}}\right) \Delta T \\ \\ \Delta T & = 67\text{$^\circ$C}\end{aligned}`

In conclusion, the temperature of the mercury metal increased by 67°C.