An ice block of 1.5 kg at -15 degrees centigrade is put in 60 kg of water at 45

Question

An ice block of 1.5 kg at -15 degrees centigrade is put in 60 kg of water at 45 degrees centigrade. What is the equilibrium temperature?

I understand that the water will lose energy and the ice will absorb that energy to first heat up the ice to 0 degrees and then melt the ice

The energy the water loses = energy the ice absorbs (both to heat up the ice and to melt it)

To warm up the ice from -15 to 0

Q=mass x spesifikk heat for ice x deltaT= 1.5kg x 2050J/kgxK 15 =46125J

Now the ice will start to melt

If I call the final temp T

46125J + heat to melt ice =heat lost from the water

46125J + 1.5kg x 334kJ/kg x K (T-T0) = 60kg x 4184(T-T0)

46125J + 1.5kg x 334kJ/kg x K (T-0) = 60kg x 4184(T-45)

Solve for T should give me the answers. However, i mess when trying to solve this and the answer make no sense. Must I use kelvin instead of degrees centigrade or am I messing up my math?

Explanation / Answer

latent heat of ice L =334 kJ/kg

specific heat of ice = 2.108 kJ/kg.K

specific heat of water = 4.187 kJ/kg.K

Heat lost by water = Heat gain by ice

- 60*4187*(T-45) = (1.5*2108*(0+15)) +(1.5*334*1000) +(1.5*4187*(T-0))

T = 41.77oC

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