Respuesta :
Answer
1.8 °C
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:
[tex]Q=mc\Delta T[/tex]To find ΔT, substitute, m = 100 g, c = 4.2 J/g•°C and Q = 750 J into the formula:
[tex]\begin{gathered} 750\text{ J }=100\text{ g }\times4.2\text{ }J\text{ /g}•\degree C\times\Delta T \\ 750\text{ J }=420\text{ J/}\degree C\times\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\times\Delta T}{420\text{ J/}\degree C} \\ \Delta T=1.7857\degree C \\ \Delta T\approx1.8\degree C \end{gathered}[/tex]The degree in °C that 100 g of water increase, if 750 J are added to it, is 1.8 °C
