calculate the amount of heat absorbed by 100 gram of ice to change into water of 30 degree Celsius (specific heat capacity of water is 1 calories per gram degree Celsius).Right answer would be marked as brainliest.
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Answered by
17
716 cal
Total heat required to convert 1 g of ice at 0°C into steam at 100°C is 716 cal.
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68
Answer:-
Given:
mass of the ice cube (m) = 100 g
Specific heat of water (s) = 1 cal / g * ° C
Original temperature(T) of ice is 0 ° C.
Temperature of water (T')= 30° C.
We know that,
Heat absorbed (Q) = ms∆T
(∆T is difference between the new temperature & original temperature.)
Hence,
⟶ Q = 100 * 1 * (30 - 0)
⟶ Q = 100 * 30
⟶ Q = 3000 cal.
⟶ Q = 3 kcal.
Therefore, the heat absorbed by the ice cube is 3 kilo calories.
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