Question

Suppose you apply a flame to heat 1 liter of water and its temperature rises by 3 C. If you apply the same flame for the same length of time to 3 liter of water, by now how much does its temperature rise

Answer 1

Answer:1$^{0}C$

Explanation:

Let the heat given by flame be $h$.

Let $m_{1}$ be the mass of water in case 1.

Let $m_{2}$ be the mass of water in case 2.

Let Δ$T_{1}$ be the temperature difference in case 1.

Let Δ$T_{2}$ be the temperature difference in case 2.

Let $V_{1}$ be the volume of water in case 1.

Let $V_{2}$ be the volume of water in case 2.

Let $d$ be the density of water.

Let $s$ be the specific heat of water.

Given,

$V_{1}=1L\\V_{2}=3L$

Δ$T_{1}=3^{0}C$

Since the same heat is given to water in both the cases,

$h=m_{1}s$Δ$T_{1}$

$h=m_{2}s$Δ$T_{2}$

So,$m_{1}s$Δ$T_{1}$=$m_{2}s$Δ$T_{2}$

Since mass is product of volume an density,

$V_{1}ds$Δ$T_{1}$=$V_{2}ds$Δ$T_{2}$

$1\times d\times s\times 3=3\times d\times s\times$Δ$T_{2}$

So,Δ$T_{2}=1^{0}C$