A $100 \mathrm{~g}$ glass container contains $250 \mathrm{~g}$ of water at $15.0^{\circ} \mathrm{C}$. A $100 \mathrm{~g}$ piece of unknown material at $100{ }^{\circ} \mathrm{C}$ is added to the water in the container. The final temperature of the mixture is $19.0^{\circ} \mathrm{C}$. What is the specific heat of the unknown material in $\mathrm{J} / \mathrm{kg}{ }^{\circ} \mathrm{C}$ ? (specific heat of water $=4,186 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C}$, specific heat of glass $=837.2$ $\mathrm{J} / \mathrm{kg}{ }^{\circ} \mathrm{C}$ )

Question

Quizplus · Accepted Answer

Explanation:Heat lost by the unknown material = Heat gained by water + Heat gained by glass container$$m_{unknown} c_{unknown} (100-19) = m_{water} c_{water} (19-15) + m_{glass} c_{glass} (19-15)$$$$m_{unknown} c_{unknown} (81) = (0.25)(4186)(4) + (0.1)(837.2)(4)$$$$m_{unknown} c_{unknown} = 418.6 + 334.88$$$$m_{unknown} c_{unknown} = 753.48$$$$c_{unknown} = \frac{753.48}{m_{unknown}}$$$$c_{unknown} = 753.48 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C}$$

A 100 g100 \mathrm{~g}100 g Glass Container Contains 250 g250 \mathrm{~g}250 g Of Water At 15.0∘C15.0^{\circ} \mathrm{C}15.0∘C

A $100 \mathrm{~g}$ Glass Container Contains $250 \mathrm{~g}$ Of Water At $15.0^{\circ} \mathrm{C}$