Answer:
Heat gained by 1.25 grams of ice is 250 joules
Explanation:
As we know
[tex]Q = mc\Delta[/tex]T
Where
[tex]\Delta[/tex]T is the change in temperature
Water freezes at 273.15 K/0 degree Celsius, and it boils at 373.15 K/00 degree Celsius,
[tex]\Delta[/tex]T
m is the mass in grams
c is the specific heat of water (ice) = 2.05 joules/gram
Substituting the given values, we get -
[tex]Q = 1.25 * 2.0 * 100\\Q = 250[/tex]Joules.
