What was the original temperature of 160.2 grams of water if placing a 31.52-g piece of aluminum with specific heat of 0.897J/g°C at 100.0°C warms the water up to 25.0°C .
I don´t get how this works, and here is my work, but it doesnt work. The answer given was 21.8°C , but i am not sure how to get that.
M x C x (Tf - Ti ) = - M x C x (Tf - Ti)
31.52g x 0.897J/g°C x (Tf-100°C) = 160.2g x 4.18J/g°C x (25°C-Ti)