Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
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.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>
mass = 12.2 kg
Explanation:
To find the mass we use the following formula (Newton's Second Law of Motion):
force = mass × acceleration
mass = force / acceleration
mass = 2.32 N / 0.19 m/s²
mass = 12.2 kg
Learn more about:
Newton's Second Law of Motion
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Answer:
Depending on the deformity it will depend on the amount of the ice that is melted,it the majority is melted that is relevant to the heated liquid melting the ice,so yes only the water formalities of the melted ice.
