Question

A liquid of mass 100g at 120°C is poured in water at 20°C,when the final temperature recorded is 40°C. If specific heat capacity of the liquid is 0.8 J/g°C,calculate the initial mass of water.
plz answer ASAP

Answer 1

Given:

A liquid of mass 100 gram at 120°C is poured in water at 20°C,when the final temperature recorded is 40°C. If specific heat capacity of the liquid is 0.8 J/g°C and specific heat capacity of the water is 4.186 J/g°C.

To Find:

Calculate the initial mass of water.

Solution:

Mass of liquid (M) = 100 g

Mass of water = m = ?

$\mathbf{\textrm{Initial temperature of liquid } = T_{i}=120^{\circ}C}$

$\mathbf{\textrm{Final temperature of liquid } = T_{f}=40^{\circ}C}$

$\mathbf{\textrm{Initial temperature of water } = t_{i}=20^{\circ}C}$

$\mathbf{\textrm{Final temperature of water} = t_{f}=40^{\circ}C}$

$\mathbf{\textrm{Specific heat capacity of liquid } = c_{pL}=0.8 \ \frac{J}{g^{\circ}C}}$

$\mathbf{\textrm{Specific heat capacity of water } = c_{pW}=4.186 \ \frac{J}{g^{\circ}C}}$

From law of conservation of energy:

Heat gain by water = Heat lost by liquid

$\mathbf{m\times c_{pW}\times (t_{f}-t_{i})=M\times c_{pL}\times (T_{i}-T_{f})}$

$\mathbf{m\times 4.186\times (40-20)=100\times 0.8\times (120-40)}$

On simplify:

m = 76.445 g

Means initial mass of water is 76.445 gram.

Answer 2

Answer:

This is the answer of the question