Answer:
Q = 30 calories
Explanation:
We have,
The specific heat of ice is 0.5 calories/gram.
The heat require to raise the temperature is given by :
[tex]Q=mc\Delta T[/tex]
m = 60 grams, c = 0.5 calories/gram °C, [tex]\Delta T=1\ C[/tex]
So,
[tex]Q=60\times 0.5\times 1\\\\Q=30\ \text{calories}[/tex]
So, 30 calories of heat is required to raise the temperature by 1 C.
