Question

A student is using a calorimeter to determine the specific heat of a metallic sample. She measures out 135.7 grams of her metal and heats it to 81.7 degrees Celsius. Then, she puts the sample into a calorimeter containing 10.82 grams of water at 48.9 degrees Celsius. She measures the temperature of the water in the calorimeter until the number stops changing, then records the final temperature to be 68.3 degrees Celsius. What is the specific heat of the metal? Please answer to three digits after the decimal point.

Answer 1

The specific heat of the metal can be calculated using the formula:

q = m × c × ΔT

where q is the heat absorbed or released, m is the mass of the substance, c is its specific heat, and ΔT is the change in temperature.

The heat absorbed by the metal is equal to the heat released by the water:

m_metal × c_metal × ΔT_metal = m_water × c_water × ΔT_water

Solving for c_metal, we get:

c_metal = (m_water × c_water × ΔT_water) / (m_metal × ΔT_metal)

Plugging in the given values, we get:

c_metal = (10.82 g × 4.184 J/g°C × (68.3 - 48.9)°C) / (135.7 g × (81.7 - 68.3)°C)

c_metal = 0.427 J/g°C (rounded to three decimal places)

Therefore, the specific heat of the metal is 0.427 J/g°C.