Question

How much heat, in joules and in calories, is required to heat 28.4g ice cube from -23 to -1

Answer 1

Given:

Mass of ice cube, m = 28.4 g

Initial temperature, T1 = -23 C

Final temperature, T2 = -1 C

To determine:

Heat required to raise the temperature from T1 to T2

Explanation:

Heat required (Q) is given as:

Q = mcΔT = mc(T2-T1)

where c = specific of ice (water) = 4.182 J/g C

Q = 28.4 * 4.182*(-1-(-23)) = 2612.9 J

Since 1 J = 0.239 cal

The corresponding Q = 0.239*2612.9 = 624.48 cal

Ans: Q = 2612.9 J

        Q = 624.4 cal

Answer 2

We have that for the Question "How much heat, in joules and in calories, is required to heat 28.4g ice cube from -23 to -1" it can be said that Heat in joules and cal is

• Q=624.4cal
• Q= 2612.9J

From the question we are told

How much heat, in joules and in calories, is required to heat 28.4g ice cube from -23 to -1

Generally the equation for the Heat Required  is mathematically given as

$Q=mcdT\\\\Q=mc(T_2-T_1)\\\\Therefore\\\\Q= 28.4 * 4.182*(-1-(-23)) \\\\Q=2612.9J$

Therefore in cal

Q=624.4cal

Hence

Heat in joules and cal is

• Q=624.4cal
• Q= 2612.9J

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