Answer :
Answer:1[tex]^{0}C[/tex]
Explanation:
Let the heat given by flame be [tex]h[/tex].
Let [tex]m_{1}[/tex] be the mass of water in case 1.
Let [tex]m_{2}[/tex] be the mass of water in case 2.
Let Δ[tex]T_{1}[/tex] be the temperature difference in case 1.
Let Δ[tex]T_{2}[/tex] be the temperature difference in case 2.
Let [tex]V_{1}[/tex] be the volume of water in case 1.
Let [tex]V_{2}[/tex] be the volume of water in case 2.
Let [tex]d[/tex] be the density of water.
Let [tex]s[/tex] be the specific heat of water.
Given,
[tex]V_{1}=1L\\V_{2}=3L\\[/tex]
Δ[tex]T_{1}=3^{0}C[/tex]
Since the same heat is given to water in both the cases,
[tex]h=m_{1}s[/tex]Δ[tex]T_{1}[/tex]
[tex]h=m_{2}s[/tex]Δ[tex]T_{2}[/tex]
So,[tex]m_{1}s[/tex]Δ[tex]T_{1}[/tex]=[tex]m_{2}s[/tex]Δ[tex]T_{2}[/tex]
Since mass is product of volume an density,
[tex]V_{1}ds[/tex]Δ[tex]T_{1}[/tex]=[tex]V_{2}ds[/tex]Δ[tex]T_{2}[/tex]
[tex]1\times d\times s\times 3=3\times d\times s\times[/tex]Δ[tex]T_{2}[/tex]
So,Δ[tex]T_{2}=1^{0}C[/tex]