My answer doesn't seem likely but I can't see where I've gone wrong:
Q: Ice of mass 0.125kg at -5 degrees is added to water of mass 0.25kg at 8 degrees. Calculate the final temperature of the water at thermal equilibrium. Assume c of ice=2100J/kg°C, c of water=4200J/kg°C and $L_f$ water=334000J/kg
My attempt: heat to turn ice to zero + heat to melt ice + heat to raise 'melted ice' to x degrees = heat released by water to x degrees.
$(0.125*5*2100)+(0.125*334000)+(0.125*4200*x)=(0.25*4200*8-x)\\1312.5+41750+525x = 1050(8-x)\\43062.5+525x=8400-1050x\\1575x=-34662.5\\x=-22.00793651\\$
I was expecting a temperature above zero.