A 0.600-kg piece of metal X is heated to $100 ^ { \circ } \mathrm { C }$ and placed in an aluminum can of mass 0.200-kg
which contains $0.500 \mathrm {~kg}$ of water initially at $17.3 ^ { \circ } \mathrm { C }$ The final equilibrium temperature of the mixture is $20.2 ^ { \circ } \mathrm { C }$ What is the specific heat of metal X? The specific heats of water and aluminum are 4186 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K }$ (water) and 910 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K } \text { (aluminum). }$

Question

A  0.600-kg piece of metal  X  is heated to $100 ^ { \circ } \mathrm { C }$ and placed in an aluminum can of mass  0.200-kg
which contains $0.500 \mathrm {~kg}$ of water initially at $17.3 ^ { \circ } \mathrm { C }$ The final equilibrium temperature of the mixture is $20.2 ^ { \circ } \mathrm { C }$ What is the specific heat of metal X? The specific heats of water and aluminum  are 4186 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K }$ (water) and 910 $\mathrm { J } / \mathrm { kg } \cdot \mathrm { K } \text { (aluminum). }$

Quizplus · Accepted Answer

The heat lost by metal X is equal to the heat gained by water and aluminum.$$Q_{lost} = Q_{gained}$$$$mC_p(T_f - T_i) = m_wC_{pw}(T_f - T_{iw}) + m_aC_{pa}(T_f - T_{ia})$$$$(0.600 	ext{ kg})C_p(20.2 ^\circ C - 100 ^\circ C) = (0.500 	ext{ kg})(4186 	ext{ J/kg$\cdot$K})(20.2 ^\circ C - 17.3 ^\circ C) + (0.200 	ext{ kg})(910 	ext{ J/kg$\cdot$K})(20.2 ^\circ C - 17.3 ^\circ C)$$$$C_p = 140 	ext{ J/kg$\cdot$K}$$

A 0600-Kg Piece of Metal X Is Heated To Heated