A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of ice is 2.10 J•g-1• °C-1. By how much will its temperature rise, assuming it does not melt? a 15,500,000 °C b 15,500 °C c 115 °C d 1.15 °C
Question
A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of ice is 2.10 J•g-1• °C-1. By how much will its temperature rise, assuming it does not melt?
a 15,500,000 °C
b 15,500 °C
c 115 °C
d 1.15 °C
Solution
To solve this problem, we need to use the formula for heat transfer which is:
q = mcΔT
where: q = heat energy (in joules) m = mass (in grams) c = specific heat capacity (in J/g°C) ΔT = change in temperature (in °C)
We are given: q = 5820 J m = 2.40 kg = 2400 g (since 1 kg = 1000 g) c = 2.10 J/g°C
We need to find ΔT. Rearranging the formula, we get:
ΔT = q / (mc)
Substituting the given values:
ΔT = 5820 J / (2400 g * 2.10 J/g°C) = 1.15 °C
So, the temperature of the ice will rise by 1.15 °C, assuming it does not melt. Therefore, the correct answer is (d) 1.15 °C.
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(d) An ice cube has a temperature of −15.0 °CThe total thermal energy needed to raise the temperature of this ice cube to 0.0 °C andcompletely melt the ice cube is 5848 Jspecific heat capacity of ice = 2100 J/kg °Cspecific latent heat of fusion of ice = 334 000 J/kgCalculate the mass of the ice cube.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Mass of ice cube = _____________________________ kg(5)(Total 9 marks)Page 3 of 34
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