Question

Using your understanding of the specific heat capacity, how many degress in °C will 100g of water (specific heat capacity =4.2J/g•°C) increase if 750J are added to it?

Answer 1

1.8 °C

Step-by-step explanation

Given data:

Mass of water, m = 100 g

Specific heat capacity of water, c = 4.2 J/g•°C

Quantity of heat added, Q = 750 J

What to find:

The degree increase in °C, ΔT

Step-by-step solution:

The formula relating specific heat and heat is given by:

`Q=mc\Delta T`

To find ΔT, substitute, m = 100 g, c = 4.2 J/g•°C and Q = 750 J into the formula:

`\begin{gathered} 750\text{ J }=100\text{ g }*4.2\text{ }J\text{ /g}•\degree C*\Delta T \\ 750\text{ J }=420\text{ J/}\degree C*\Delta T \\ \text{Divide both side by 420 J/}\degree C \\ \frac{750\text{ J}}{420\text{ J /}\degree C}=\frac{420\text{ J/}\degree C*\Delta T}{420\text{ J/}\degree C} \\ \Delta T=1.7857\degree C \\ \Delta T\approx1.8\degree C \end{gathered}`

The degree in °C that 100 g of water increase, if 750 J are added to it, is 1.8 °C