Question

1. what is the temperature change of 250g of water that has 1980j of heat added to it? (the specific heat of water is 4.18 j/g°c.)

Explanation:

Step1: Recall the heat - transfer formula

The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat added (or removed), \(m\) is the mass of the substance, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. We need to solve for \(\Delta T\). Rearranging the formula gives \(\Delta T=\frac{Q}{mc}\).

Step2: Substitute the given values

We are given that \(Q = 1980\space J\), \(m=250\space g\), and \(c = 4.18\space J/g^{\circ}C\).
Substitute these values into the formula: \(\Delta T=\frac{1980}{250\times4.18}\).
First, calculate the denominator: \(250\times4.18 = 1045\).
Then, \(\Delta T=\frac{1980}{1045}\approx1.895^{\circ}C\).

Answer:

The temperature change \(\Delta T\approx1.9^{\circ}C\)