Question

A 458 mL of water at 21.5°C is in a calorimeter. If a gold at 99.7°C is in put into the calorimeter, the calorimeter becomes 25.3°C. Specific heat of gold= 0.129 J/g-K

How many grams of gold was put into the calorimeter?

How much heat was involved in the processes?

Answer 1

$m_g \times c_g \times (T_g -T_e) = m_w \times c_w \times (T_e -T_w) \\ m_g \times 0.129 \times (99.7 -25.3) = 458 \times 4.184 \times (25.3 -21.5) \\ m_g \times 9.5976 = 7281.8336 \\ m_g = \frac{7281.8336}{9.5976} = 758.71 \; g \\ Q = m_w c_w (T_e -T_w) \\ Q = 458 \times 4.184 \times (25.3-21.5) = 7281.83 \;J$