Recent content by mtu8

Temp. of the rod when all ice has frozen: -196 + 14.56 + 77.685 = -103.755 degrees The temp. of the ice at this point would be 0 degrees celsius right? The system is not in thermal equilibrium at this point, so the rod should continue to cool the ice past zero degrees. I just don't know...

Q (15-->0) = (.0265)(4186)(15) = 1663.9 J 1663.9 J = (.175)(653) ∆T *the 653 is the specific heat of aluminum over this temp. range ∆T (15-->0) = 14.56 Q (freeze) = (.0265)(33.5e4) *the 33.5e4 is the latent heat of fusion Q (freeze) = 8877.5 J 8877.5 J = (.175)(653)∆T ∆T...

Conservation of energy problem! HELP PLEASE! 1. Homework Statement A 175 g aluminum cylinder is removed from a liquid nitrogen bath, where it has been cooled to -196°C. The cylinder is immediately placed in an insulated cup containing 26.5 g of water at 15.0°C, and the system is allowed to...

m(a) * c(a) * (T(f) - T(i)) = m(i) * c(i) * (T(f) - T(i)) (.175 kg) (653) (T(f) + 196) = (.0265) (2100) (T(f) - 0)) *I'm sure I'm just not getting the algebra right or something...

is this right.. Q (al) = (.175)(653)(196) = 22,397.9 J

Total Heat = (8877.5) + (1663.9) = 10541.4 J

How do you find the total amount of heat needed to freeze the water? Q (water to freeze) = (.0265)(33.5e4) = 8877.5 J

Here are my calculations, any help as to what I’m doing wrong would be appreciated. Q (15 --> 0) = (.0265)(4186)(15) = 1663.9 J 1663.9 J = (.175)(653) ∆T ∆T (15 --> 0) = 14.56 Q (freeze) = (.0265)(33.5e4) Q (freeze) = 8877.5 J 8877.5 J = (.175)(653) ∆T ∆T (freeze) =...

Homework Statement A 175 g aluminum cylinder is removed from a liquid nitrogen bath, where it has been cooled to -196°C. The cylinder is immediately placed in an insulated cup containing 26.5 g of water at 15.0°C, and the system is allowed to come to equilibrium. Determine if all of the water...