Question

1. if 335 g of water at 65.5 ^{circ}c loses 9750 j of heat, what is the final temperature of the water? liquid water has a specific heat of 4.18 j/(gcdot^{circ}c).

Explanation:

Step1: Recall the heat - transfer formula

The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat lost or gained, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\). Since heat is lost, \(Q=- 9750\space J\), \(m = 335\space g\), \(c = 4.18\space J/(g\cdot^{\circ}C)\), and \(T_{initial}=65.5^{\circ}C\).

Step2: Rearrange the formula to solve for \(\Delta T\)

From \(Q = mc\Delta T\), we can solve for \(\Delta T\) as \(\Delta T=\frac{Q}{mc}\).
Substitute the values: \(\Delta T=\frac{-9750\space J}{335\space g\times4.18\space J/(g\cdot^{\circ}C)}\)

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Step3: Solve for \(T_{final}\)

Since \(\Delta T=T_{final}-T_{initial}\), then \(T_{final}=\Delta T + T_{initial}\)
Substitute \(\Delta T=-6.96^{\circ}C\) and \(T_{initial} = 65.5^{\circ}C\)
\(T_{final}=-6.96^{\circ}C + 65.5^{\circ}C\)
\(T_{final}=58.54^{\circ}C\approx58.5^{\circ}C\)

Answer:

\(58.5^{\circ}C\)