cool the 100 g soup 10 degrees from 50 to 40
warm the x g water 35 deg from 5 degrees to 40 deg
heat out = .8 * 100 * 10
heat in = 1.0 * x * 35
so
35 x = 800
x = 22.8 g
which is 23 to your rough significant figures
A hostess wants to cool her 100g of soup which is currently at 50Celsius, and it's specific heat is 0.8 cal/g celsius, by adding water (specific heat = 1 cal/ g celsius) at a temperature of 5Celsius, how much water should she add, assuming that all the heat from the soup goes into the water (and it must also end up at 40Celsius)
the answer is mass=23 g I just don't know what equations to use and how to find the answer
1 answer
1 answer