Question

Note: specific heat capacity of water C= 4184 J/kg 0 CWith 31,840 Joules, a water was heated raising its temperature from 22.0 °C to 28.5 °C. Find the mass of the water, in grams. Q =m=ΔT=C=Solution:

Answer 1

We are given the following information

Specific heat capacity of water = 4184 J/kg.°C

Amount of heat = 31,840 Joules

Change in Temperature = 22.0 °C to 28.5 °C

We are asked to find the mass of the water in grams.

Recall the specific heat capacity formula is given by

$Q=m\cdot c\cdot\Delta T$

Let us substitute the given values and solve for mass (m)

$\begin{gathered} Q=m\cdot c\cdot\Delta T \\ 31,840=m\cdot4184\cdot(28.5-22.0) \\ 31,840=m\cdot4184\cdot6.5 \\ 31,840=m\cdot27,196 \\ m=\frac{31,840}{27,196} \\ m=1.1708\; kg \end{gathered}$

Multiply the mass in kg by 1000 to convert into grams.

$m=1.1708\times1000=1170.8\; g$

Therefore, the mass of the water is 1170.8 grams.