Question

A 200-gram liquid sample of Alcohol X is prepared at -10°C. The sample is then added to 300 g of water at 20°C in a sealed styrofoam container. When thermal equilibrium is reached, the temperature of the alcohol-water solution is 10°C. What is the specific heat capacity of the alcohol? Assume the sealed container is an isolated system. The specific heat capacity of water is 4.19 kJ/kg · °C.

Answer 1

The attachment shows my working. This can be done by using the formula for mixtures —>

Energy lost = energy gained. In this case, the energy lost by the alcohol is gained by the water.
Sooo, the formula for heat energy is mcΔt,
So you go
McΔt (alcohol) = mcΔt (water)
And then algebraically solve for C.

Refer to my working for the answer. Cheers!!