When 4256.4 Joules of heat is absorbed, a 150 gram sample of a metal warms from 25.1 degrees Celsius to 65.8 degrees Celsius .What is the specific dot (C)p=(Q)/(m(Delta T)) heat (Cp) of the metal?The formula for calculating specific heat has been given. 0.70(J)/(gcdot ^circ )C 0.6971(J)/(gcdot ^circ )C 0.697(J)/(gcdot ^circ )C 0.7(J)/(gcdot ^circ )C

To find the specific heat (\(C_p\)) of the metal, we can use the formula provided:<br /><br />\[ C_p = \frac{Q}{m \Delta T} \]<br /><br />where:<br />- \( Q \) is the amount of heat absorbed (4256.4 Joules),<br />- \( m \) is the mass of the sample (150 grams),<br />- \( \Delta T \) is the change in temperature.<br /><br />First, calculate the change in temperature (\(\Delta T\)):<br /><br />\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \]<br />\[ \Delta T = 65.8^\circ C - 25.1^\circ C \]<br />\[ \Delta T = 40.7^\circ C \]<br /><br />Now, substitute the values into the formula:<br /><br />\[ C_p = \frac{4256.4 \, \text{J}}{150 \, \text{g} \times 40.7^\circ C} \]<br /><br />Calculate the denominator:<br /><br />\[ 150 \, \text{g} \times 40.7^\circ C = 6105 \, \text{g} \cdot ^\circ C \]<br /><br />Now, divide the numerator by the denominator:<br /><br />\[ C_p = \frac{4256.4 \, \text{J}}{6105 \, \text{g} \cdot ^\circ C} \]<br />\[ C_p \approx 0.6971 \, \frac{\text{J}}{\text{g} \cdot ^\circ C} \]<br /><br />Therefore, the specific heat (\(C_p\)) of the metal is approximately:<br /><br />\[ 0.6971 \, \frac{\text{J}}{\text{g} \cdot ^\circ C} \]<br /><br />The correct answer is:<br /><br />\[ 0.6971 \, \frac{\text{J}}{\text{g} \cdot ^\circ C} \]