1. A mass of 0.35kg of ice at −15∘C is lowered into an insulated beaker containing 0.61kg of water at 59∘C. What is the temperature after equilibrium has been reached? Give your answer in ∘C
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Hopefully, you have solved the problem. Just in case you still need help, see the explanation below.
This is a slightly tricky question. There are a few quantities that you need to compute first before you can set up the conservation of energy to solve for the equilibrium temperature.
Amount of thermal energy need to raise the ice from −15°C to 0 °C is
Amount of thermal energy need to melt all the ice is
Amount of thermal energy released when the water from 59°C to 0 °C is
Since Q3 > Q1 + Q2, the equilibrium temperature would be above 0 °C but less than 59°C because there is enough thermal energy released in cooling the water from 59°C to the equilibrium temperature, Teq to raise the ice from −15°C to 0 °C and melt all the ice.
Assume there is no heat loss and consider the system to be the ice and hot water, conservation of energy would give the following:
micecice(0 – (−15)) + miceLfusion + mice cwater (Teq – 0) + mwatercwater(Teq – 59) =0
I would leave the algebra of solving for Teq to you.