Question

How much heat is needed to bring 25.5 g of water from 29.3 °C to 43.87 °C.Q =m=ΔT= C=solution =

Answer 1

We are given the following information

Mass of water = 25.5 g

Initial temperature of water = 29.3 °C

Final temperature of water = 43.87 °C

The specific heat capacity of water is 4.186 J/g.°C

The amount of heat required is given by

`\begin{gathered} Q=m\cdot c\cdot\Delta T \\ Q=m\cdot c\cdot(T_f-T_i) \end{gathered}`

Let us substitute the given values into the above formula

`\begin{gathered} Q=25.5*4.186\cdot(43.87-29.3) \\ Q=1,555.25\; J \end{gathered}`

Therefore, we need 1,555.25 Joules of heat to bring 25.5 g of water from 29.3 °C to 43.87 °C.