Answer:
The temperature difference is [tex]\frac{300}{m} Kkg^{-1}[/tex]
Explanation:
The formula that is to used is :
Δ[tex]Q=ms[/tex]Δ[tex]T[/tex]
where ,
- Δ[tex]Q[/tex] is the heat supplied in calories = 300cal
- [tex]m[/tex] is the mass of water taken = m (assumed)
- Δ[tex]T[/tex] is the change in temperature
- [tex]s[/tex] is the specific heat of water = [tex]1calK^{-1}Kg^{-1}[/tex]
ΔT :
[tex]\frac{300}{m*1} \\=\frac{300}{m} Kkg^{-1}[/tex]
