Question

It takes 2,267 joules of heat to raise the temperature of a 44.5 gram sample of a metal from 33.9°C to 288.3°C. What is the heat capacity of the metal?

Answer 1

The heat capacity of the metal is approximately 3.55 J/g°C.

Step-by-step explanation:

The heat capacity of a substance is the amount of heat required to raise the temperature of a given amount of the substance by one degree Celsius. It is represented by the symbol 'C'. To calculate the heat capacity of the metal, we can use the formula:

C = Q / (m ΔT)

where C is the heat capacity, Q is the heat absorbed, m is the mass of the metal, and ΔT is the change in temperature.

In this case, the heat absorbed is 2,267 joules, the mass of the metal is 44.5 grams, and the change in temperature is 288.3°C - 33.9°C = 254.4°C.

Substituting these values into the formula, we can calculate:

C = 2,267 J / (44.5 g × 254.4°C)

After performing the calculations, the heat capacity of the metal is approximately 3.55 J/g°C.

Answer 2

`\boxed{\text{0.200 J$^(\circ)$C$^(-1)$g$^(-1)$}}}`

Step-by-step explanation:

The formula for the amount heat q absorbed by a substance is

q = mcΔT

where

m = the mass of the substance

C = the specific heat capacity of the material

ΔT = the temperature change

Data:

q = 2267 J

m = 44.5 g

T₁ = 33.9 °C

T₂ = 288.3 °C

Calculations:

ΔT = (288.3 - 33.9) °C = 254.4 °C

`\begin{array}{rcl}2267 & = & 44.5 * C * 254.4\\2267 & = & 11 180C\\C& = & (2267)/(11180)\\\\& = & \textbf{0.200 J$^(\circ)$C$^(-1)$g$^(-1)$}\\\end{array}`

`\text{The specific heat capacity of the metal is \boxed{\textbf{0.200 J$^(\circ)$C$^(-1)$g$^(-1)$}}}`