an 11.5 kg object with a temperature of 25.0 degrees celsius is attached to a pulley mechanism. it is allowed to fall until is slows to rest after falling 1.6 m. at the end of the fall , the object has a temperature of 28.9 degrees celsius. what is the specific heat of the object?
assuming the lost potential energy converts to heat in the object,
PEchang=mass*specheat*deltatemp
11.5*9.8*1.6=11.5*specheat*3.9
solve for specific heat