Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Step-by-step explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
Q = m.c.ΔT,
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
It is known that 1.0 cal = 4.18 J.
∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.