Question

formula: $q = mcdelta t$

1. how much heat must be absorbed by 375 grams of water to raise its temperature by $25^{circ}c$? (the specific heat of water is $4.18 j/g^{circ}c$.)

Explanation:

Step1: Identify the values

Given \(m = 375\space g\), \(c=4.18\space J/g^{\circ}C\), \(\Delta T = 25^{\circ}C\)

Step2: Substitute into the formula

Use the formula \(Q=mc\Delta T\). Substitute the values: \(Q=(375\space g)\times(4.18\space J/g^{\circ}C)\times(25^{\circ}C)\)

Step3: Calculate the result

First, multiply \(375\times4.18 = 1567.5\). Then \(1567.5\times25=39187.5\space J\)

Answer:

\(39187.5\space J\)