Answer:1500 calorie
Explanation:
Given
volume of water [tex]V=50\ ml[/tex]
Initial temperature [tex]T_i=22^{\circ}C[/tex]
Final temperature [tex]T_f=52^{\circ}C[/tex]
We know,
specific heat of water is [tex]c=1\ calorie/gm-^{\circ}C[/tex]
the density of water is [tex]\rho=1\ gm/ml[/tex]
The heat required to raise the temperature
[tex]\Rightarrow Q=mc\Delta T\\\Rightarrow Q=\rho \cdot V\times c\times \Delta T\\\Rightarrow Q=1\times 50\times 1\times (52-22)\\\Rightarrow Q=1500\ cal.[/tex]
1500 calories is required
